(20+2x)(30+2x)=1440

Solution for (20+2x)(30+2x)=1440 equation:

(20+2x)(30+2x)=1440

We move all terms to the left:

(20+2x)(30+2x)-(1440)=0

We add all the numbers together, and all the variables

(2x+20)(2x+30)-1440=0

We multiply parentheses ..

(+4x^2+60x+40x+600)-1440=0

We get rid of parentheses

4x^2+60x+40x+600-1440=0

We add all the numbers together, and all the variables

4x^2+100x-840=0

a = 4; b = 100; c = -840;
Δ = b2-4ac
Δ = 1002-4·4·(-840)
Δ = 23440
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:

$\sqrt{\Delta}=\sqrt{23440}=\sqrt{16*1465}=\sqrt{16}*\sqrt{1465}=4\sqrt{1465}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-4\sqrt{1465}}{2*4}=\frac{-100-4\sqrt{1465}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+4\sqrt{1465}}{2*4}=\frac{-100+4\sqrt{1465}}{8}
$