How do you find the roots, real and imaginary, of $y = - 2 x^{2} + 7 x + 6$ $y=-2x^2 + 7x +6$ using the quadratic formula?

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Answer 1 · 2015-12-01T17:38:41

Plug in a, b, c in the quadratic equation which is $\frac{(- b) \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $((-b)+-sqrt(b^2-4ac))/(2a)$

$a =$ $a=$ coefficient of $x^{2}$ $x^2$ .
$a = - 2$ $a=-2$

$b =$ $b=$ coefficient of $x$ $x$ .
$b = 7$ $b=7$

$c =$ $c=$ the constant in the equation.
$c = 6$ $c=6$

$\frac{(- 7) \pm \sqrt{{(7)}^{2} - 4 (- 2) (6)}}{2 (- 2)}$ $((-7)+-sqrt((7)^2-4(-2)(6)))/(2(-2))$

How do you find the roots, real and imaginary, of y=-2x^2 + 7x +6 y=−2x2+7x+6y=-2x^2 + 7x +6 using the quadratic formula?