4x(12-x)=40

Solution for 4x(12-x)=40 equation:

4x(12-x)=40

We move all terms to the left:

4x(12-x)-(40)=0

We add all the numbers together, and all the variables

4x(-1x+12)-40=0

We multiply parentheses

-4x^2+48x-40=0

a = -4; b = 48; c = -40;
Δ = b2-4ac
Δ = 482-4·(-4)·(-40)
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{26}}{2*-4}=\frac{-48-8\sqrt{26}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{26}}{2*-4}=\frac{-48+8\sqrt{26}}{-8}
$