(8x-3)(2+x)+(8x-3)(x-7)=0

Solution for (8x-3)(2+x)+(8x-3)(x-7)=0 equation:

(8x-3)(2+x)+(8x-3)(x-7)=0

We add all the numbers together, and all the variables

(8x-3)(x+2)+(8x-3)(x-7)=0

We multiply parentheses ..

(+8x^2+16x-3x-6)+(8x-3)(x-7)=0

We get rid of parentheses

8x^2+16x-3x+(8x-3)(x-7)-6=0

We multiply parentheses ..

8x^2+(+8x^2-56x-3x+21)+16x-3x-6=0

We add all the numbers together, and all the variables

8x^2+(+8x^2-56x-3x+21)+13x-6=0

We get rid of parentheses

8x^2+8x^2-56x-3x+13x+21-6=0

We add all the numbers together, and all the variables

16x^2-46x+15=0

a = 16; b = -46; c = +15;
Δ = b2-4ac
Δ = -462-4·16·15
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-34}{2*16}=\frac{12}{32}
=3/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+34}{2*16}=\frac{80}{32}
=2+1/2
$