(x^2+2)-8(3+x)=-34

Solution for (x^2+2)-8(3+x)=-34 equation:

(x^2+2)-8(3+x)=-34

We move all terms to the left:

(x^2+2)-8(3+x)-(-34)=0

We add all the numbers together, and all the variables

(x^2+2)-8(x+3)-(-34)=0

We add all the numbers together, and all the variables

(x^2+2)-8(x+3)+34=0

We multiply parentheses

(x^2+2)-8x-24+34=0

We get rid of parentheses

x^2-8x+2-24+34=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 = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$