(36+2x)(48+2x)=1900

Solution for (36+2x)(48+2x)=1900 equation:

(36+2x)(48+2x)=1900

We move all terms to the left:

(36+2x)(48+2x)-(1900)=0

We add all the numbers together, and all the variables

(2x+36)(2x+48)-1900=0

We multiply parentheses ..

(+4x^2+96x+72x+1728)-1900=0

We get rid of parentheses

4x^2+96x+72x+1728-1900=0

We add all the numbers together, and all the variables

4x^2+168x-172=0

a = 4; b = 168; c = -172;
Δ = b2-4ac
Δ = 1682-4·4·(-172)
Δ = 30976
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{30976}=176$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(168)-176}{2*4}=\frac{-344}{8}
=-43
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(168)+176}{2*4}=\frac{8}{8}
=1
$