n^2+23n-4=32

Solution for n^2+23n-4=32 equation:

n^2+23n-4=32

We move all terms to the left:

n^2+23n-4-(32)=0

We add all the numbers together, and all the variables

n^2+23n-36=0

a = 1; b = 23; c = -36;
Δ = b2-4ac
Δ = 232-4·1·(-36)
Δ = 673
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{673}}{2*1}=\frac{-23-\sqrt{673}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{673}}{2*1}=\frac{-23+\sqrt{673}}{2}
$