Answer :
To solve the quadratic equation [tex]\( 6x^2 - x - 15 = 0 \)[/tex], follow these steps:
1. Identify the coefficients: In the given quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are [tex]\( a = 6 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -15 \)[/tex].
2. Calculate the discriminant: The discriminant ([tex]\( \Delta \)[/tex]) for a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
3. Plug the coefficients into the discriminant formula:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 6 \cdot (-15) = 1 + 360 = 361 \][/tex]
4. Find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{361} = 19 \][/tex]
5. Apply the quadratic formula: The solutions for [tex]\( x \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
6. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\(\sqrt{\Delta} \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm 19}{2 \cdot 6} = \frac{1 \pm 19}{12} \][/tex]
7. Calculate the two solutions:
- For the positive case ([tex]\( + \)[/tex]):
[tex]\[ x_1 = \frac{1 + 19}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.6666666666666667 \][/tex]
- For the negative case ([tex]\( - \)[/tex]):
[tex]\[ x_2 = \frac{1 - 19}{12} = \frac{-18}{12} = -1.5 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( 6x^2 - x - 15 = 0 \)[/tex] are:
[tex]\[ x = \frac{5}{3} \approx 1.6666666666666667, \, -1.5 \][/tex]
1. Identify the coefficients: In the given quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are [tex]\( a = 6 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -15 \)[/tex].
2. Calculate the discriminant: The discriminant ([tex]\( \Delta \)[/tex]) for a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
3. Plug the coefficients into the discriminant formula:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 6 \cdot (-15) = 1 + 360 = 361 \][/tex]
4. Find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{361} = 19 \][/tex]
5. Apply the quadratic formula: The solutions for [tex]\( x \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
6. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\(\sqrt{\Delta} \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm 19}{2 \cdot 6} = \frac{1 \pm 19}{12} \][/tex]
7. Calculate the two solutions:
- For the positive case ([tex]\( + \)[/tex]):
[tex]\[ x_1 = \frac{1 + 19}{12} = \frac{20}{12} = \frac{5}{3} \approx 1.6666666666666667 \][/tex]
- For the negative case ([tex]\( - \)[/tex]):
[tex]\[ x_2 = \frac{1 - 19}{12} = \frac{-18}{12} = -1.5 \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( 6x^2 - x - 15 = 0 \)[/tex] are:
[tex]\[ x = \frac{5}{3} \approx 1.6666666666666667, \, -1.5 \][/tex]
