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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +6;
Δ = b2-4ac
Δ = 32-4·(-1)·6
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*-1}=\frac{-3-\sqrt{33}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*-1}=\frac{-3+\sqrt{33}}{-2}
$