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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+8x-3=0

a = -3; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·(-3)·(-3)
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{7}}{2*-3}=\frac{-8-2\sqrt{7}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{7}}{2*-3}=\frac{-8+2\sqrt{7}}{-6}
$