12k^2+20k-9=0

Solution for 12k^2+20k-9=0 equation:

12k^2+20k-9=0

a = 12; b = 20; c = -9;
Δ = b2-4ac
Δ = 202-4·12·(-9)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{13}}{2*12}=\frac{-20-8\sqrt{13}}{24}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{13}}{2*12}=\frac{-20+8\sqrt{13}}{24}
$