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x^2+12x+18=9
We move all terms to the left:
x^2+12x+18-(9)=0
We add all the numbers together, and all the variables
x^2+12x+9=0
a = 1; b = 12; c = +9;
Δ = b2-4ac
Δ = 122-4·1·9
Δ = 108
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{3}}{2*1}=\frac{-12-6\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{3}}{2*1}=\frac{-12+6\sqrt{3}}{2} $
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