# How do you solve the equation: 4x^2+13x-75=0?

Solve $y = 4 {x}^{2} + 13 x - 75 = 0$ (1)
Transformed equation $y ' = {x}^{2} + 13 x - 300 = 0$ (2). Roots have different signs.
Factor pairs of ac = -300 ->.... (-10, 30)(-12, 25). This sum is 13 = b. Then, the 2 real roots of (2) are: y1 = 12 and y2 = -25. Back to original equation (1), the 2 real roots are: $x 1 = \frac{y 1}{a} = \frac{12}{4} = 3$, and
$x 2 = \frac{y 2}{a} = - \frac{25}{4.}$.