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

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

(2-x2)-(3x^2=4x)-(-1-2x)

We move all terms to the left:

(2-x2)-(3x^2-(4x)-(-1-2x))=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


We calculate terms in parentheses: -(3x^2-4x-(-2x-1)), so:

3x^2-4x-(-2x-1)

We get rid of parentheses

3x^2-4x+2x+1

We add all the numbers together, and all the variables

3x^2-2x+1

Back to the equation:

-(3x^2-2x+1)


We get rid of parentheses

-1x^2-3x^2+2x-1+2=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}
$