20x^2+20x-75=0

Solution for 20x^2+20x-75=0 equation:

20x^2+20x-75=0

a = 20; b = 20; c = -75;
Δ = b2-4ac
Δ = 202-4·20·(-75)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-80}{2*20}=\frac{-100}{40}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+80}{2*20}=\frac{60}{40}
=1+1/2
$