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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(8(x+2)), so:

8(x+2)

We multiply parentheses

8x+16

Back to the equation:

-(8x+16)


We get rid of parentheses

x^2-8x-16+4=0

We add all the numbers together, and all the variables

x^2-8x-12=0

a = 1; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·1·(-12)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{7}}{2*1}=\frac{8-4\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*1}=\frac{8+4\sqrt{7}}{2}
$