9k^2+12k-23=0

Solution for 9k^2+12k-23=0 equation:

9k^2+12k-23=0

a = 9; b = 12; c = -23;
Δ = b2-4ac
Δ = 122-4·9·(-23)
Δ = 972
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:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18\sqrt{3}}{2*9}=\frac{-12-18\sqrt{3}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18\sqrt{3}}{2*9}=\frac{-12+18\sqrt{3}}{18}
$