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

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

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

We move all terms to the left:

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-1x^2+6x-9+1=0

We add all the numbers together, and all the variables

-1x^2+6x-8=0

a = -1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·(-1)·(-8)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-1}=\frac{-4}{-2}
=+2
$