x^2+16x+48=8

Solution for x^2+16x+48=8 equation:

x^2+16x+48=8

We move all terms to the left:

x^2+16x+48-(8)=0

We add all the numbers together, and all the variables

x^2+16x+40=0

a = 1; b = 16; c = +40;
Δ = b2-4ac
Δ = 162-4·1·40
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{6}}{2*1}=\frac{-16-4\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{6}}{2*1}=\frac{-16+4\sqrt{6}}{2}
$