(7+2x)(5+2x)-35=35

Solution for (7+2x)(5+2x)-35=35 equation:

(7+2x)(5+2x)-35=35

We move all terms to the left:

(7+2x)(5+2x)-35-(35)=0

We add all the numbers together, and all the variables

(2x+7)(2x+5)-35-35=0

We add all the numbers together, and all the variables

(2x+7)(2x+5)-70=0

We multiply parentheses ..

(+4x^2+10x+14x+35)-70=0

We get rid of parentheses

4x^2+10x+14x+35-70=0

We add all the numbers together, and all the variables

4x^2+24x-35=0

a = 4; b = 24; c = -35;
Δ = b2-4ac
Δ = 242-4·4·(-35)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{71}}{2*4}=\frac{-24-4\sqrt{71}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{71}}{2*4}=\frac{-24+4\sqrt{71}}{8}
$