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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(8+3x(x+4))+64=0


We calculate terms in parentheses: -(8+3x(x+4)), so:

8+3x(x+4)

determiningTheFunctionDomain
3x(x+4)+8

We multiply parentheses

3x^2+12x+8

Back to the equation:

-(3x^2+12x+8)


We get rid of parentheses

-3x^2-12x-8+64=0

We add all the numbers together, and all the variables

-3x^2-12x+56=0

a = -3; b = -12; c = +56;
Δ = b2-4ac
Δ = -122-4·(-3)·56
Δ = 816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{51}}{2*-3}=\frac{12-4\sqrt{51}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{51}}{2*-3}=\frac{12+4\sqrt{51}}{-6}
$