How do you find the zeros, real and imaginary, of $y=-5x^2-13x-4$ using the quadratic formula?

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Answer 1 · 2017-07-09T02:15:56

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 $a$ ; $-13$ for $b$ and $-4$ for $c$ gives:

$x = (-(-13) +- sqrt((-13)^2 - (4 * -5 * -4)))/(2 * -5)$

$x = (13 +- sqrt(169 - 80))/(-10)$

$x = (13 +- sqrt(89))/(-10)$

$x = -13/10 +- -sqrt(89)/10$

$x = -13/10 + sqrt(89)/10$ and $x = -13/10 - sqrt(89)/10$

How do you find the zeros, real and imaginary, of y=-5x^2-13x-4y=-5x^2-13x-4 using the quadratic formula?