4u(u-3)+8(2u-3)=4

Solution for 4u(u-3)+8(2u-3)=4 equation:

4u(u-3)+8(2u-3)=4

We move all terms to the left:

4u(u-3)+8(2u-3)-(4)=0

We multiply parentheses

4u^2-12u+16u-24-4=0

We add all the numbers together, and all the variables

4u^2+4u-28=0

a = 4; b = 4; c = -28;
Δ = b2-4ac
Δ = 42-4·4·(-28)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{29}}{2*4}=\frac{-4-4\sqrt{29}}{8}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{29}}{2*4}=\frac{-4+4\sqrt{29}}{8}
$