x+(2x^2+100x)/(3x+100)=100

Solution for x+(2x^2+100x)/(3x+100)=100 equation:

x+(2x^2+100x)/(3x+100)=100

We move all terms to the left:

x+(2x^2+100x)/(3x+100)-(100)=0


Domain of the equation: (3x+100)!=0

We move all terms containing x to the left, all other terms to the right

3x!=-100

x!=-100/3

x!=-33+1/3

x∈R


We multiply all the terms by the denominator


x*(3x+100)+(2x^2+100x)-100*(3x+100)=0

We multiply parentheses

3x^2+100x+(2x^2+100x)-300x-10000=0

We get rid of parentheses

3x^2+2x^2+100x+100x-300x-10000=0

We add all the numbers together, and all the variables

5x^2-100x-10000=0

a = 5; b = -100; c = -10000;
Δ = b2-4ac
Δ = -1002-4·5·(-10000)
Δ = 210000
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:

$\sqrt{\Delta}=\sqrt{210000}=\sqrt{10000*21}=\sqrt{10000}*\sqrt{21}=100\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-100\sqrt{21}}{2*5}=\frac{100-100\sqrt{21}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+100\sqrt{21}}{2*5}=\frac{100+100\sqrt{21}}{10}
$