(x^2-6x+3)=(3x+2x^2-1)

Solution for (x^2-6x+3)=(3x+2x^2-1) equation:

(x^2-6x+3)=(3x+2x^2-1)

We move all terms to the left:

(x^2-6x+3)-((3x+2x^2-1))=0

We get rid of parentheses

-((3x+2x^2-1))+x^2-6x+3=0


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

(3x+2x^2-1)

We get rid of parentheses

2x^2+3x-1

Back to the equation:

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


We add all the numbers together, and all the variables

x^2-6x-(2x^2+3x-1)+3=0

We get rid of parentheses

x^2-2x^2-6x-3x+1+3=0

We add all the numbers together, and all the variables

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

a = -1; b = -9; c = +4;
Δ = b2-4ac
Δ = -92-4·(-1)·4
Δ = 97
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{-(-9)-\sqrt{97}}{2*-1}=\frac{9-\sqrt{97}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{97}}{2*-1}=\frac{9+\sqrt{97}}{-2}
$