n^2-23n+120=0

Solution for n^2-23n+120=0 equation:

n^2-23n+120=0

a = 1; b = -23; c = +120;
Δ = b2-4ac
Δ = -232-4·1·120
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*1}=\frac{16}{2}
=8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*1}=\frac{30}{2}
=15
$