How do you find the zeros, real and imaginary, of $y=-5x^2-2x-9$ using the quadratic formula?

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Answer 1 · 2017-06-09T02:57:37

See a solution process below:

For $ax^2 + bx + c = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

Substituting $-5$ for $a$ ; $-2$ for $b$ and $-9$ for $c$ gives:

$x = (-(-2) +- sqrt((-2)^2 - (4 * -5 * -9)))/(2 * -5)$

$x = (2 +- sqrt(4 - 180))/(-10)$

$x = (2 +- sqrt(-176))/(-10)$

$x = (2 +- sqrt(16 * -11))/(-10)$

$x = (2 +- (sqrt(16) * sqrt(-11)))/(-10)$

$x = (2 +- 4sqrt(-11))/(-10)$

$x = -(1 +- 2sqrt(-11))/5$

How do you find the zeros, real and imaginary, of y=-5x^2-2x-9y=-5x^2-2x-9 using the quadratic formula?