u^2+(4u/5)=5

Solution for u^2+(4u/5)=5 equation:

u^2+(4u/5)=5

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

u^2+4u/5-5=0

We multiply all the terms by the denominator


u^2*5+4u-5*5=0

We add all the numbers together, and all the variables

u^2*5+4u-25=0

Wy multiply elements

5u^2+4u-25=0

a = 5; b = 4; c = -25;
Δ = b2-4ac
Δ = 42-4·5·(-25)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{129}}{2*5}=\frac{-4-2\sqrt{129}}{10}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{129}}{2*5}=\frac{-4+2\sqrt{129}}{10}
$