2x^2+4X=3+3x

Solution for 2x^2+4X=3+3x equation:

2x^2+4x=3+3x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+x-3=0

a = 2; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*2}=\frac{4}{4}
=1
$