12k^2+32k+8=0

Solution for 12k^2+32k+8=0 equation:

12k^2+32k+8=0

a = 12; b = 32; c = +8;
Δ = b2-4ac
Δ = 322-4·12·8
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{10}}{2*12}=\frac{-32-8\sqrt{10}}{24}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{10}}{2*12}=\frac{-32+8\sqrt{10}}{24}
$