8x(4x-3)+8=x+5(x+1)

Solution for 8x(4x-3)+8=x+5(x+1) equation:

8x(4x-3)+8=x+5(x+1)

We move all terms to the left:

8x(4x-3)+8-(x+5(x+1))=0

We multiply parentheses

32x^2-24x-(x+5(x+1))+8=0


We calculate terms in parentheses: -(x+5(x+1)), so:

x+5(x+1)

We multiply parentheses

x+5x+5

We add all the numbers together, and all the variables

6x+5

Back to the equation:

-(6x+5)


We get rid of parentheses

32x^2-24x-6x-5+8=0

We add all the numbers together, and all the variables

32x^2-30x+3=0

a = 32; b = -30; c = +3;
Δ = b2-4ac
Δ = -302-4·32·3
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{129}}{2*32}=\frac{30-2\sqrt{129}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{129}}{2*32}=\frac{30+2\sqrt{129}}{64}
$