s^2+23s-324=0

Solution for s^2+23s-324=0 equation:

s^2+23s-324=0

a = 1; b = 23; c = -324;
Δ = b2-4ac
Δ = 232-4·1·(-324)
Δ = 1825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-5\sqrt{73}}{2*1}=\frac{-23-5\sqrt{73}}{2}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+5\sqrt{73}}{2*1}=\frac{-23+5\sqrt{73}}{2}
$