(-2x^2+6x+1)-2=(4x^2-3x+1)

Solution for (-2x^2+6x+1)-2=(4x^2-3x+1) equation:

(-2x^2+6x+1)-2=(4x^2-3x+1)

We move all terms to the left:

(-2x^2+6x+1)-2-((4x^2-3x+1))=0

We get rid of parentheses

-2x^2+6x-((4x^2-3x+1))+1-2=0


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

(4x^2-3x+1)

We get rid of parentheses

4x^2-3x+1

Back to the equation:

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


We add all the numbers together, and all the variables

-2x^2+6x-(4x^2-3x+1)-1=0

We get rid of parentheses

-2x^2-4x^2+6x+3x-1-1=0

We add all the numbers together, and all the variables

-6x^2+9x-2=0

a = -6; b = 9; c = -2;
Δ = b2-4ac
Δ = 92-4·(-6)·(-2)
Δ = 33
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{33}}{2*-6}=\frac{-9-\sqrt{33}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{33}}{2*-6}=\frac{-9+\sqrt{33}}{-12}
$