(35+2x)(93+2x)=4059

Solution for (35+2x)(93+2x)=4059 equation:

(35+2x)(93+2x)=4059

We move all terms to the left:

(35+2x)(93+2x)-(4059)=0

We add all the numbers together, and all the variables

(2x+35)(2x+93)-4059=0

We multiply parentheses ..

(+4x^2+186x+70x+3255)-4059=0

We get rid of parentheses

4x^2+186x+70x+3255-4059=0

We add all the numbers together, and all the variables

4x^2+256x-804=0

a = 4; b = 256; c = -804;
Δ = b2-4ac
Δ = 2562-4·4·(-804)
Δ = 78400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{78400}=280$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(256)-280}{2*4}=\frac{-536}{8}
=-67
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(256)+280}{2*4}=\frac{24}{8}
=3
$