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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+5x-6=0

a = -1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·(-1)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-1}=\frac{-6}{-2}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-1}=\frac{-4}{-2}
=+2
$