u(u-5)+8u=u(u+2)4

Solution for u(u-5)+8u=u(u+2)4 equation:

u(u-5)+8u=u(u+2)4

We move all terms to the left:

u(u-5)+8u-(u(u+2)4)=0

We add all the numbers together, and all the variables

8u+u(u-5)-(u(u+2)4)=0

We multiply parentheses

u^2+8u-5u-(u(u+2)4)=0


We calculate terms in parentheses: -(u(u+2)4), so:

u(u+2)4

We multiply parentheses

4u^2+8u

Back to the equation:

-(4u^2+8u)


We add all the numbers together, and all the variables

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

We get rid of parentheses

u^2-4u^2+3u-8u=0

We add all the numbers together, and all the variables

-3u^2-5u=0

a = -3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-3)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-3}=\frac{0}{-6}
=0
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-3}=\frac{10}{-6}
=-1+2/3
$