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

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Answer 1 · 2017-07-16T00:05:10

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 $a$ ; $1$ for $b$ and $-4$ for $c$ gives:

$x = (-1 +- sqrt(1^2 - (4 * -5 * -4)))/(2 * -5)$

$x = (-1 +- sqrt(1 - (80)))/(-10)$

$x = (-1 +- sqrt(-79))/(-10)$

$x = (-1 + sqrt(-79))/(-10)$ and $x = (-1 - sqrt(-79))/(-10)$

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