0=n2+23n-150

Solution for 0=n2+23n-150 equation:

0=n2+23n-150

We move all terms to the left:

0-(n2+23n-150)=0

We add all the numbers together, and all the variables

-(+n^2+23n-150)+0=0

We add all the numbers together, and all the variables

-(+n^2+23n-150)=0

We get rid of parentheses

-n^2-23n+150=0

We add all the numbers together, and all the variables

-1n^2-23n+150=0

a = -1; b = -23; c = +150;
Δ = b2-4ac
Δ = -232-4·(-1)·150
Δ = 1129
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{1129}}{2*-1}=\frac{23-\sqrt{1129}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1129}}{2*-1}=\frac{23+\sqrt{1129}}{-2}
$