80=(2x+5)*(x-10)

Solution for 80=(2x+5)*(x-10) equation:

80=(2x+5)(x-10)

We move all terms to the left:

80-((2x+5)(x-10))=0

We multiply parentheses ..

-((+2x^2-20x+5x-50))+80=0


We calculate terms in parentheses: -((+2x^2-20x+5x-50)), so:

(+2x^2-20x+5x-50)

We get rid of parentheses

2x^2-20x+5x-50

We add all the numbers together, and all the variables

2x^2-15x-50

Back to the equation:

-(2x^2-15x-50)


We get rid of parentheses

-2x^2+15x+50+80=0

We add all the numbers together, and all the variables

-2x^2+15x+130=0

a = -2; b = 15; c = +130;
Δ = b2-4ac
Δ = 152-4·(-2)·130
Δ = 1265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{1265}}{2*-2}=\frac{-15-\sqrt{1265}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{1265}}{2*-2}=\frac{-15+\sqrt{1265}}{-4}
$