3x+4x+8=x^2

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

3x+4x+8=x^2

We move all terms to the left:

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

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

We add all the numbers together, and all the variables

-1x^2+7x+8=0

a = -1; b = 7; c = +8;
Δ = b2-4ac
Δ = 72-4·(-1)·8
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-1}=\frac{2}{-2}
=-1
$