(2-x)2+(3x+1)(3x-1)-4=5x2

Solution for (2-x)2+(3x+1)(3x-1)-4=5x2 equation:

(2-x)2+(3x+1)(3x-1)-4=5x^2

We move all terms to the left:

(2-x)2+(3x+1)(3x-1)-4-(5x^2)=0

determiningTheFunctionDomain
-5x^2+(2-x)2+(3x+1)(3x-1)-4=0

We add all the numbers together, and all the variables

-5x^2+(-1x+2)2+(3x+1)(3x-1)-4=0

We use the square of the difference formula

-5x^2+9x^2+(-1x+2)2-1-4=0

We multiply parentheses

-5x^2+9x^2-2x+4-1-4=0

We add all the numbers together, and all the variables

4x^2-2x-1=0

a = 4; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·4·(-1)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*4}=\frac{2-2\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*4}=\frac{2+2\sqrt{5}}{8}
$