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3x+6=x^2+6x+8
We move all terms to the left:
3x+6-(x^2+6x+8)=0
We get rid of parentheses
-x^2+3x-6x-8+6=0
We add all the numbers together, and all the variables
-1x^2-3x-2=0
a = -1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·(-1)·(-2)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-1}=\frac{2}{-2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-1}=\frac{4}{-2} =-2 $
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