8x2+32x-4=20

Solution for 8x2+32x-4=20 equation:

8x^2+32x-4=20

We move all terms to the left:

8x^2+32x-4-(20)=0

We add all the numbers together, and all the variables

8x^2+32x-24=0

a = 8; b = 32; c = -24;
Δ = b2-4ac
Δ = 322-4·8·(-24)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{7}}{2*8}=\frac{-32-16\sqrt{7}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{7}}{2*8}=\frac{-32+16\sqrt{7}}{16}
$