80=(x-3)(x-5)

Solution for 80=(x-3)(x-5) equation:

80=(x-3)(x-5)

We move all terms to the left:

80-((x-3)(x-5))=0

We multiply parentheses ..

-((+x^2-5x-3x+15))+80=0


We calculate terms in parentheses: -((+x^2-5x-3x+15)), so:

(+x^2-5x-3x+15)

We get rid of parentheses

x^2-5x-3x+15

We add all the numbers together, and all the variables

x^2-8x+15

Back to the equation:

-(x^2-8x+15)


We get rid of parentheses

-x^2+8x-15+80=0

We add all the numbers together, and all the variables

-1x^2+8x+65=0

a = -1; b = 8; c = +65;
Δ = b2-4ac
Δ = 82-4·(-1)·65
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-18}{2*-1}=\frac{-26}{-2}
=+13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+18}{2*-1}=\frac{10}{-2}
=-5
$