(x2-2x)2+3(x2-2x)-4=0

Solution for (x2-2x)2+3(x2-2x)-4=0 equation:

(x2-2x)2+3(x2-2x)-4=0

We add all the numbers together, and all the variables

(+x^2-2x)2+3(+x^2-2x)-4=0

We multiply parentheses

2x^2+3x^2-4x-6x-4=0

We add all the numbers together, and all the variables

5x^2-10x-4=0

a = 5; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·5·(-4)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{5}}{2*5}=\frac{10-6\sqrt{5}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{5}}{2*5}=\frac{10+6\sqrt{5}}{10}
$