120+80x=13x^2+39x

Solution for 120+80x=13x^2+39x equation:

120+80x=13x^2+39x

We move all terms to the left:

120+80x-(13x^2+39x)=0

We get rid of parentheses

-13x^2+80x-39x+120=0

We add all the numbers together, and all the variables

-13x^2+41x+120=0

a = -13; b = 41; c = +120;
Δ = b2-4ac
Δ = 412-4·(-13)·120
Δ = 7921
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7921}=89$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-89}{2*-13}=\frac{-130}{-26}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+89}{2*-13}=\frac{48}{-26}
=-1+11/13
$