To solve the quadratic equation [tex]\(2x^2 + x - 5 = 0\)[/tex] using the quadratic formula, we follow a series of steps. The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are:
- [tex]\(a = 2\)[/tex]
- [tex]\(b = 1\)[/tex]
- [tex]\(c = -5\)[/tex]
### Step 1: Calculate the discriminant
The discriminant [tex]\(\Delta\)[/tex] is calculated using the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 1^2 - 4(2)(-5) \][/tex]
[tex]\[ \Delta = 1 - (-40) \][/tex]
[tex]\[ \Delta = 1 + 40 \][/tex]
[tex]\[ \Delta = 41 \][/tex]
So, the discriminant is 41.
### Step 2: Calculate the roots using the quadratic formula
Using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(\Delta = 41\)[/tex], [tex]\(a = 2\)[/tex], and [tex]\(b = 1\)[/tex]:
[tex]\[ x = \frac{-1 \pm \sqrt{41}}{2(2)} \][/tex]
[tex]\[ x = \frac{-1 \pm \sqrt{41}}{4} \][/tex]
This expression gives us two solutions, one for the plus sign ( [tex]\(+\)[/tex] ) and one for the minus sign ( [tex]\(-\)[/tex] ).
### Step 3: Compute the two solutions
First solution ([tex]\(+\)[/tex]):
[tex]\[ x_1 = \frac{-1 + \sqrt{41}}{4} \][/tex]
[tex]\[ x_1 \approx \frac{-1 + 6.403124237}{4} \][/tex]
[tex]\[ x_1 \approx \frac{5.403124237}{4} \][/tex]
[tex]\[ x_1 \approx 1.3507810593582121 \][/tex]
Second solution ([tex]\(-\)[/tex]):
[tex]\[ x_2 = \frac{-1 - \sqrt{41}}{4} \][/tex]
[tex]\[ x_2 \approx \frac{-1 - 6.403124237}{4} \][/tex]
[tex]\[ x_2 \approx \frac{-7.403124237}{4} \][/tex]
[tex]\[ x2 \approx -1.8507810593582121 \][/tex]
### Summary:
The solutions to the quadratic equation [tex]\(2x^2 + x - 5 = 0\)[/tex] are:
[tex]\[ x_1 \approx 1.3507810593582121 \][/tex]
[tex]\[ x_2 \approx -1.8507810593582121 \][/tex]
And the discriminant is 41.
