12k^2+4k-7=0

Solution for 12k^2+4k-7=0 equation:

12k^2+4k-7=0

a = 12; b = 4; c = -7;
Δ = b2-4ac
Δ = 42-4·12·(-7)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{22}}{2*12}=\frac{-4-4\sqrt{22}}{24}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{22}}{2*12}=\frac{-4+4\sqrt{22}}{24}
$