3(2+x)=3x2+4

Solution for 3(2+x)=3x2+4 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-3)·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{-(3)-\sqrt{33}}{2*-3}=\frac{-3-\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*-3}=\frac{-3+\sqrt{33}}{-6}
$