n^2-4n=10n+15

Solution for n^2-4n=10n+15 equation:

n^2-4n=10n+15

We move all terms to the left:

n^2-4n-(10n+15)=0

We get rid of parentheses

n^2-4n-10n-15=0

We add all the numbers together, and all the variables

n^2-14n-15=0

a = 1; b = -14; c = -15;
Δ = b2-4ac
Δ = -142-4·1·(-15)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*1}=\frac{-2}{2}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*1}=\frac{30}{2}
=15
$