4x-8+3x-4=x^2

Solution for 4x-8+3x-4=x^2 equation:

4x-8+3x-4=x^2

We move all terms to the left:

4x-8+3x-4-(x^2)=0

determiningTheFunctionDomain
-x^2+4x+3x-8-4=0

We add all the numbers together, and all the variables

-1x^2+7x-12=0

a = -1; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·(-1)·(-12)
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{-(7)-1}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*-1}=\frac{-6}{-2}
=+3
$