(2x+16)(4x+3)=35

Solution for (2x+16)(4x+3)=35 equation:

(2x+16)(4x+3)=35

We move all terms to the left:

(2x+16)(4x+3)-(35)=0

We multiply parentheses ..

(+8x^2+6x+64x+48)-35=0

We get rid of parentheses

8x^2+6x+64x+48-35=0

We add all the numbers together, and all the variables

8x^2+70x+13=0

a = 8; b = 70; c = +13;
Δ = b2-4ac
Δ = 702-4·8·13
Δ = 4484
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:

$\sqrt{\Delta}=\sqrt{4484}=\sqrt{4*1121}=\sqrt{4}*\sqrt{1121}=2\sqrt{1121}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1121}}{2*8}=\frac{-70-2\sqrt{1121}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1121}}{2*8}=\frac{-70+2\sqrt{1121}}{16}
$