(4x+3)(x-8)+(x-1)(4x+3)=0

Solution for (4x+3)(x-8)+(x-1)(4x+3)=0 equation:

(4x+3)(x-8)+(x-1)(4x+3)=0

We multiply parentheses ..

(+4x^2-32x+3x-24)+(x-1)(4x+3)=0

We get rid of parentheses

4x^2-32x+3x+(x-1)(4x+3)-24=0

We multiply parentheses ..

4x^2+(+4x^2+3x-4x-3)-32x+3x-24=0

We add all the numbers together, and all the variables

4x^2+(+4x^2+3x-4x-3)-29x-24=0

We get rid of parentheses

4x^2+4x^2+3x-4x-29x-3-24=0

We add all the numbers together, and all the variables

8x^2-30x-27=0

a = 8; b = -30; c = -27;
Δ = b2-4ac
Δ = -302-4·8·(-27)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-42}{2*8}=\frac{-12}{16}
=-3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+42}{2*8}=\frac{72}{16}
=4+1/2
$