Answer :
Sure, I'll guide you through solving both equations step-by-step.
### Equation 1:
[tex]\[ x^2 - 10 = -5x \][/tex]
Step 1: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ x^2 - 10 + 5x = 0 \][/tex]
or
[tex]\[ x^2 + 5x - 10 = 0 \][/tex]
Step 2: This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ a = 1, \quad b = 5, \quad c = -10 \][/tex]
Step 3: Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to find the roots.
[tex]\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times -10}}{2 \times 1} \][/tex]
Step 4: Solve under the square root:
[tex]\[ x = \frac{-5 \pm \sqrt{25 + 40}}{2} \][/tex]
[tex]\[ x = \frac{-5 \pm \sqrt{65}}{2} \][/tex]
Step 5: Write the final solutions:
[tex]\[ x_1 = \frac{-5 + \sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = \frac{-5 - \sqrt{65}}{2} \][/tex]
So, the solutions are:
[tex]\[ x_1 = -\frac{5}{2} + \frac{\sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{65}}{2} - \frac{5}{2} \][/tex]
---
### Equation 2:
[tex]\[ 2(x^2 - x) = 6 \][/tex]
Step 1: Distribute the 2 on the left side.
[tex]\[ 2x^2 - 2x = 6 \][/tex]
Step 2: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ 2x^2 - 2x - 6 = 0 \][/tex]
Step 3: This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ a = 2, \quad b = -2, \quad c = -6 \][/tex]
Step 4: Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to find the roots.
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times -6}}{2 \times 2} \][/tex]
Step 5: Solve under the square root:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 48}}{4} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{52}}{4} \][/tex]
Step 6: Simplify the square root of 52:
[tex]\[ \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \][/tex]
Step 7: Substitute back:
[tex]\[ x = \frac{2 \pm 2\sqrt{13}}{4} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{13}}{2} \][/tex]
Step 8: Write the final solutions:
[tex]\[ x_1 = \frac{1 - \sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1 + \sqrt{13}}{2} \][/tex]
So, the solutions are:
[tex]\[ x_1 = \frac{1}{2} - \frac{\sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1}{2} + \frac{\sqrt{13}}{2} \][/tex]
### Conclusion:
For equation [tex]\( x^2 - 10 = -5x \)[/tex]:
[tex]\[ x_1 = -\frac{5}{2} + \frac{\sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{65}}{2} - \frac{5}{2} \][/tex]
For equation [tex]\( 2(x^2 - x) = 6 \)[/tex]:
[tex]\[ x_1 = \frac{1}{2} - \frac{\sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1}{2} + \frac{\sqrt{13}}{2} \][/tex]
### Equation 1:
[tex]\[ x^2 - 10 = -5x \][/tex]
Step 1: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ x^2 - 10 + 5x = 0 \][/tex]
or
[tex]\[ x^2 + 5x - 10 = 0 \][/tex]
Step 2: This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ a = 1, \quad b = 5, \quad c = -10 \][/tex]
Step 3: Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to find the roots.
[tex]\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \times 1 \times -10}}{2 \times 1} \][/tex]
Step 4: Solve under the square root:
[tex]\[ x = \frac{-5 \pm \sqrt{25 + 40}}{2} \][/tex]
[tex]\[ x = \frac{-5 \pm \sqrt{65}}{2} \][/tex]
Step 5: Write the final solutions:
[tex]\[ x_1 = \frac{-5 + \sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = \frac{-5 - \sqrt{65}}{2} \][/tex]
So, the solutions are:
[tex]\[ x_1 = -\frac{5}{2} + \frac{\sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{65}}{2} - \frac{5}{2} \][/tex]
---
### Equation 2:
[tex]\[ 2(x^2 - x) = 6 \][/tex]
Step 1: Distribute the 2 on the left side.
[tex]\[ 2x^2 - 2x = 6 \][/tex]
Step 2: Move all terms to one side of the equation to set it equal to zero.
[tex]\[ 2x^2 - 2x - 6 = 0 \][/tex]
Step 3: This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. Identify [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ a = 2, \quad b = -2, \quad c = -6 \][/tex]
Step 4: Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to find the roots.
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times -6}}{2 \times 2} \][/tex]
Step 5: Solve under the square root:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 48}}{4} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{52}}{4} \][/tex]
Step 6: Simplify the square root of 52:
[tex]\[ \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \][/tex]
Step 7: Substitute back:
[tex]\[ x = \frac{2 \pm 2\sqrt{13}}{4} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{13}}{2} \][/tex]
Step 8: Write the final solutions:
[tex]\[ x_1 = \frac{1 - \sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1 + \sqrt{13}}{2} \][/tex]
So, the solutions are:
[tex]\[ x_1 = \frac{1}{2} - \frac{\sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1}{2} + \frac{\sqrt{13}}{2} \][/tex]
### Conclusion:
For equation [tex]\( x^2 - 10 = -5x \)[/tex]:
[tex]\[ x_1 = -\frac{5}{2} + \frac{\sqrt{65}}{2} \][/tex]
[tex]\[ x_2 = -\frac{\sqrt{65}}{2} - \frac{5}{2} \][/tex]
For equation [tex]\( 2(x^2 - x) = 6 \)[/tex]:
[tex]\[ x_1 = \frac{1}{2} - \frac{\sqrt{13}}{2} \][/tex]
[tex]\[ x_2 = \frac{1}{2} + \frac{\sqrt{13}}{2} \][/tex]
