20x^2+16x-75=0

Solution for 20x^2+16x-75=0 equation:

20x^2+16x-75=0

a = 20; b = 16; c = -75;
Δ = b2-4ac
Δ = 162-4·20·(-75)
Δ = 6256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6256}=\sqrt{16*391}=\sqrt{16}*\sqrt{391}=4\sqrt{391}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{391}}{2*20}=\frac{-16-4\sqrt{391}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{391}}{2*20}=\frac{-16+4\sqrt{391}}{40}
$