128=(x^2)/(x+5)

Solution for 128=(x^2)/(x+5) equation:

128=(x^2)/(x+5)

We move all terms to the left:

128-((x^2)/(x+5))=0


Domain of the equation: (x+5))!=0

x∈R


We multiply all the terms by the denominator


-(x^2+128*(x+5))=0


We calculate terms in parentheses: -(x^2+128*(x+5)), so:

x^2+128*(x+5)

We multiply parentheses

x^2+128x+640

Back to the equation:

-(x^2+128x+640)


We get rid of parentheses

-x^2-128x-640=0

We add all the numbers together, and all the variables

-1x^2-128x-640=0

a = -1; b = -128; c = -640;
Δ = b2-4ac
Δ = -1282-4·(-1)·(-640)
Δ = 13824
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{13824}=\sqrt{2304*6}=\sqrt{2304}*\sqrt{6}=48\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-48\sqrt{6}}{2*-1}=\frac{128-48\sqrt{6}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+48\sqrt{6}}{2*-1}=\frac{128+48\sqrt{6}}{-2}
$