3u+8u(u-8)=-20

Solution for 3u+8u(u-8)=-20 equation:

3u+8u(u-8)=-20

We move all terms to the left:

3u+8u(u-8)-(-20)=0

We add all the numbers together, and all the variables

3u+8u(u-8)+20=0

We multiply parentheses

8u^2+3u-64u+20=0

We add all the numbers together, and all the variables

8u^2-61u+20=0

a = 8; b = -61; c = +20;
Δ = b2-4ac
Δ = -612-4·8·20
Δ = 3081
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-61)-\sqrt{3081}}{2*8}=\frac{61-\sqrt{3081}}{16}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-61)+\sqrt{3081}}{2*8}=\frac{61+\sqrt{3081}}{16}
$