n2-19n+88=0

Solution for n2-19n+88=0 equation:

n2-19n+88=0

We add all the numbers together, and all the variables

n^2-19n+88=0

a = 1; b = -19; c = +88;
Δ = b2-4ac
Δ = -192-4·1·88
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-3}{2*1}=\frac{16}{2}
=8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+3}{2*1}=\frac{22}{2}
=11
$