(30+2x)(50+2x)=1900

Solution for (30+2x)(50+2x)=1900 equation:

(30+2x)(50+2x)=1900

We move all terms to the left:

(30+2x)(50+2x)-(1900)=0

We add all the numbers together, and all the variables

(2x+30)(2x+50)-1900=0

We multiply parentheses ..

(+4x^2+100x+60x+1500)-1900=0

We get rid of parentheses

4x^2+100x+60x+1500-1900=0

We add all the numbers together, and all the variables

4x^2+160x-400=0

a = 4; b = 160; c = -400;
Δ = b2-4ac
Δ = 1602-4·4·(-400)
Δ = 32000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32000}=\sqrt{6400*5}=\sqrt{6400}*\sqrt{5}=80\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-80\sqrt{5}}{2*4}=\frac{-160-80\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+80\sqrt{5}}{2*4}=\frac{-160+80\sqrt{5}}{8}
$