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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

(2x+x^2-1)

We get rid of parentheses

x^2+2x-1

Back to the equation:

-(x^2+2x-1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2-5x+1=0

a = 3; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·3·1
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{13}}{2*3}=\frac{5-\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{13}}{2*3}=\frac{5+\sqrt{13}}{6}
$