(14+2x)(20+2x)=616

Solution for (14+2x)(20+2x)=616 equation:

(14+2x)(20+2x)=616

We move all terms to the left:

(14+2x)(20+2x)-(616)=0

We add all the numbers together, and all the variables

(2x+14)(2x+20)-616=0

We multiply parentheses ..

(+4x^2+40x+28x+280)-616=0

We get rid of parentheses

4x^2+40x+28x+280-616=0

We add all the numbers together, and all the variables

4x^2+68x-336=0

a = 4; b = 68; c = -336;
Δ = b2-4ac
Δ = 682-4·4·(-336)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-100}{2*4}=\frac{-168}{8}
=-21
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+100}{2*4}=\frac{32}{8}
=4
$