25n+110=12(n+4)25n+110=12(n+4)

Solution for 25n+110=12(n+4)25n+110=12(n+4) equation:

25n+110=12(n+4)25n+110=12(n+4)

We move all terms to the left:

25n+110-(12(n+4)25n+110)=0


We calculate terms in parentheses: -(12(n+4)25n+110), so:

12(n+4)25n+110

We multiply parentheses

300n^2+1200n+110

Back to the equation:

-(300n^2+1200n+110)


We get rid of parentheses

-300n^2+25n-1200n-110+110=0

We add all the numbers together, and all the variables

-300n^2-1175n=0

a = -300; b = -1175; c = 0;
Δ = b2-4ac
Δ = -11752-4·(-300)·0
Δ = 1380625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1380625}=1175$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1175)-1175}{2*-300}=\frac{0}{-600}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1175)+1175}{2*-300}=\frac{2350}{-600}
=-3+11/12
$