2x^2/(5+5x-10x^2)=2

Solution for 2x^2/(5+5x-10x^2)=2 equation:

2x^2/(5+5x-10x^2)=2

We move all terms to the left:

2x^2/(5+5x-10x^2)-(2)=0


Domain of the equation: (5+5x-10x^2)!=0

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

-10x^2+5x!=-5

x∈R


We multiply all the terms by the denominator


2x^2-2*(5+5x-10x^2)=0

We multiply parentheses

2x^2+20x^2-10x-10=0

We add all the numbers together, and all the variables

22x^2-10x-10=0

a = 22; b = -10; c = -10;
Δ = b2-4ac
Δ = -102-4·22·(-10)
Δ = 980
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:

$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14\sqrt{5}}{2*22}=\frac{10-14\sqrt{5}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14\sqrt{5}}{2*22}=\frac{10+14\sqrt{5}}{44}
$