20x^2+15x-575=0

Solution for 20x^2+15x-575=0 equation:

20x^2+15x-575=0

a = 20; b = 15; c = -575;
Δ = b2-4ac
Δ = 152-4·20·(-575)
Δ = 46225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{46225}=215$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-215}{2*20}=\frac{-230}{40}
=-5+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+215}{2*20}=\frac{200}{40}
=5
$