(25+2x)(20+2x)-500=306

Solution for (25+2x)(20+2x)-500=306 equation:

(25+2x)(20+2x)-500=306

We move all terms to the left:

(25+2x)(20+2x)-500-(306)=0

We add all the numbers together, and all the variables

(2x+25)(2x+20)-500-306=0

We add all the numbers together, and all the variables

(2x+25)(2x+20)-806=0

We multiply parentheses ..

(+4x^2+40x+50x+500)-806=0

We get rid of parentheses

4x^2+40x+50x+500-806=0

We add all the numbers together, and all the variables

4x^2+90x-306=0

a = 4; b = 90; c = -306;
Δ = b2-4ac
Δ = 902-4·4·(-306)
Δ = 12996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12996}=114$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-114}{2*4}=\frac{-204}{8}
=-25+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+114}{2*4}=\frac{24}{8}
=3
$