How do you find the roots, real and imaginary, of $y=-15^2 + 12x +34$ using the quadratic formula?

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Answer 1 · 2017-07-19T14:12:00

See a solution process below:

(Assuming the problem is: $y = -15x^2 + 12x + 34$ )

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

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

Substituting $-15$ for $a$ ; $12$ for $b$ and $34$ for $c$ gives:

$x = (-12 +- sqrt(12^2 - (4 * -15 * 34)))/(2 * -15)$

$x = (-12 +- sqrt(144 - (-2040)))/(-30)$

$x = (-12 +- sqrt(144 + 2040))/(-30)$

$x = (-12 +- sqrt(2184))/(-30)$

$x = (-12)/(-30) +- (sqrt(2184))/(-30)$

$x = 2/5 +- (sqrt(2184))/(-30)$

$x = 2/5 +- (sqrt(4 * 546))/(-30)$

$x = 2/5 +- (sqrt(4)sqrt(546))/(-30)$

$x = 2/5 +- (2sqrt(546))/(-30)$

$x = 2/5 +- (sqrt(546))/(-15)$

$x = 2/5 - (sqrt(546))/(15)$ and $x = 2/5 + (sqrt(546))/(15)$

How do you find the roots, real and imaginary, of y=-15^2 + 12x +34 y=-15^2 + 12x +34 using the quadratic formula?