k^2+24k-15=0

Solution for k^2+24k-15=0 equation:

k^2+24k-15=0

a = 1; b = 24; c = -15;
Δ = b2-4ac
Δ = 242-4·1·(-15)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{159}}{2*1}=\frac{-24-2\sqrt{159}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{159}}{2*1}=\frac{-24+2\sqrt{159}}{2}
$