4x(6x-10)=8x+40

Solution for 4x(6x-10)=8x+40 equation:

4x(6x-10)=8x+40

We move all terms to the left:

4x(6x-10)-(8x+40)=0

We multiply parentheses

24x^2-40x-(8x+40)=0

We get rid of parentheses

24x^2-40x-8x-40=0

We add all the numbers together, and all the variables

24x^2-48x-40=0

a = 24; b = -48; c = -40;
Δ = b2-4ac
Δ = -482-4·24·(-40)
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32\sqrt{6}}{2*24}=\frac{48-32\sqrt{6}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32\sqrt{6}}{2*24}=\frac{48+32\sqrt{6}}{48}
$