8i^2+68i+32=0

Solution for 8i^2+68i+32=0 equation:

8i^2+68i+32=0

a = 8; b = 68; c = +32;
Δ = b2-4ac
Δ = 682-4·8·32
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-60}{2*8}=\frac{-128}{16}
=-8
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+60}{2*8}=\frac{-8}{16}
=-1/2
$