(5)/(x^2+2x-1)=2

Solution for (5)/(x^2+2x-1)=2 equation:

(5)/(x^2+2x-1)=2

We move all terms to the left:

(5)/(x^2+2x-1)-(2)=0


Domain of the equation: (x^2+2x-1)!=0

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

x^2+2x!=1

x∈R


We multiply all the terms by the denominator


-2*(x^2+2x-1)+5=0

We multiply parentheses

-2x^2-4x+2+5=0

We add all the numbers together, and all the variables

-2x^2-4x+7=0

a = -2; b = -4; c = +7;
Δ = b2-4ac
Δ = -42-4·(-2)·7
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{2}}{2*-2}=\frac{4-6\sqrt{2}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{2}}{2*-2}=\frac{4+6\sqrt{2}}{-4}
$