8x(x-5)=8x+40

Solution for 8x(x-5)=8x+40 equation:

8x(x-5)=8x+40

We move all terms to the left:

8x(x-5)-(8x+40)=0

We multiply parentheses

8x^2-40x-(8x+40)=0

We get rid of parentheses

8x^2-40x-8x-40=0

We add all the numbers together, and all the variables

8x^2-48x-40=0

a = 8; b = -48; c = -40;
Δ = b2-4ac
Δ = -482-4·8·(-40)
Δ = 3584
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:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{14}}{2*8}=\frac{48-16\sqrt{14}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{14}}{2*8}=\frac{48+16\sqrt{14}}{16}
$