8a^2+32a-4=0

Solution for 8a^2+32a-4=0 equation:

8a^2+32a-4=0

a = 8; b = 32; c = -4;
Δ = b2-4ac
Δ = 322-4·8·(-4)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-24\sqrt{2}}{2*8}=\frac{-32-24\sqrt{2}}{16}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+24\sqrt{2}}{2*8}=\frac{-32+24\sqrt{2}}{16}
$