(x-3)=x*(2x+4)-80

Solution for (x-3)=x*(2x+4)-80 equation:

(x-3)=x(2x+4)-80

We move all terms to the left:

(x-3)-(x(2x+4)-80)=0

We get rid of parentheses

x-(x(2x+4)-80)-3=0


We calculate terms in parentheses: -(x(2x+4)-80), so:

x(2x+4)-80

We multiply parentheses

2x^2+4x-80

Back to the equation:

-(2x^2+4x-80)


We get rid of parentheses

-2x^2+x-4x+80-3=0

We add all the numbers together, and all the variables

-2x^2-3x+77=0

a = -2; b = -3; c = +77;
Δ = b2-4ac
Δ = -32-4·(-2)·77
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-25}{2*-2}=\frac{-22}{-4}
=5+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+25}{2*-2}=\frac{28}{-4}
=-7
$