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To solve the quadratic equation [tex]\(-2x^2 + 11x + 5 = 0\)[/tex], let's follow these steps:
1. Identify the coefficients: In the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], we have:
[tex]\[ a = -2, \quad b = 11, \quad c = 5 \][/tex]
2. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
The discriminant is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 11^2 - 4(-2)(5) = 121 + 40 = 161 \][/tex]
3. Calculate the roots using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-11 \pm \sqrt{161}}{2(-2)} = \frac{-11 \pm \sqrt{161}}{-4} \][/tex]
4. Simplify the roots:
Simplifying the expression for the roots:
[tex]\[ x_1 = \frac{11 - \sqrt{161}}{4} \][/tex]
[tex]\[ x_2 = \frac{11 + \sqrt{161}}{4} \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(-2x^2 + 11x + 5 = 0\)[/tex] are:
[tex]\[ x_1 = \frac{11 - \sqrt{161}}{4}, \quad x_2 = \frac{11 + \sqrt{161}}{4} \][/tex]
1. Identify the coefficients: In the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], we have:
[tex]\[ a = -2, \quad b = 11, \quad c = 5 \][/tex]
2. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
The discriminant is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 11^2 - 4(-2)(5) = 121 + 40 = 161 \][/tex]
3. Calculate the roots using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-11 \pm \sqrt{161}}{2(-2)} = \frac{-11 \pm \sqrt{161}}{-4} \][/tex]
4. Simplify the roots:
Simplifying the expression for the roots:
[tex]\[ x_1 = \frac{11 - \sqrt{161}}{4} \][/tex]
[tex]\[ x_2 = \frac{11 + \sqrt{161}}{4} \][/tex]
Therefore, the solutions to the quadratic equation [tex]\(-2x^2 + 11x + 5 = 0\)[/tex] are:
[tex]\[ x_1 = \frac{11 - \sqrt{161}}{4}, \quad x_2 = \frac{11 + \sqrt{161}}{4} \][/tex]
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