0=(80+2x)(60+2x)-4800

Solution for 0=(80+2x)(60+2x)-4800 equation:

0=(80+2x)(60+2x)-4800

We move all terms to the left:

0-((80+2x)(60+2x)-4800)=0

We add all the numbers together, and all the variables

-((2x+80)(2x+60)-4800)+0=0

We add all the numbers together, and all the variables

-((2x+80)(2x+60)-4800)=0

We multiply parentheses ..

-((+4x^2+120x+160x+4800)-4800)=0


We calculate terms in parentheses: -((+4x^2+120x+160x+4800)-4800), so:

(+4x^2+120x+160x+4800)-4800

We get rid of parentheses

4x^2+120x+160x+4800-4800

We add all the numbers together, and all the variables

4x^2+280x

Back to the equation:

-(4x^2+280x)


We get rid of parentheses

-4x^2-280x=0

a = -4; b = -280; c = 0;
Δ = b2-4ac
Δ = -2802-4·(-4)·0
Δ = 78400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{78400}=280$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-280)-280}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-280)+280}{2*-4}=\frac{560}{-8}
=-70
$