x^2+8x+8x+45=-6x

Solution for x^2+8x+8x+45=-6x equation:

x^2+8x+8x+45=-6x

We move all terms to the left:

x^2+8x+8x+45-(-6x)=0

We add all the numbers together, and all the variables

x^2+16x-(-6x)+45=0

We get rid of parentheses

x^2+16x+6x+45=0

We add all the numbers together, and all the variables

x^2+22x+45=0

a = 1; b = 22; c = +45;
Δ = b2-4ac
Δ = 222-4·1·45
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-4\sqrt{19}}{2*1}=\frac{-22-4\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+4\sqrt{19}}{2*1}=\frac{-22+4\sqrt{19}}{2}
$