12k^2+4k-8=0

Solution for 12k^2+4k-8=0 equation:

12k^2+4k-8=0

a = 12; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·12·(-8)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*12}=\frac{-24}{24}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*12}=\frac{16}{24}
=2/3
$