4u-16(12-u)=(12-u)u

Solution for 4u-16(12-u)=(12-u)u equation:

4u-16(12-u)=(12-u)u

We move all terms to the left:

4u-16(12-u)-((12-u)u)=0

We add all the numbers together, and all the variables

4u-16(-1u+12)-((-1u+12)u)=0

We multiply parentheses

4u+16u-((-1u+12)u)-192=0


We calculate terms in parentheses: -((-1u+12)u), so:

(-1u+12)u

We multiply parentheses

-1u^2+12u

Back to the equation:

-(-1u^2+12u)


We add all the numbers together, and all the variables

-(-1u^2+12u)+20u-192=0

We get rid of parentheses

1u^2-12u+20u-192=0

We add all the numbers together, and all the variables

u^2+8u-192=0

a = 1; b = 8; c = -192;
Δ = b2-4ac
Δ = 82-4·1·(-192)
Δ = 832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{13}}{2*1}=\frac{-8-8\sqrt{13}}{2}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{13}}{2*1}=\frac{-8+8\sqrt{13}}{2}
$