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0=x^2+12x+12
We move all terms to the left:
0-(x^2+12x+12)=0
We add all the numbers together, and all the variables
-(x^2+12x+12)=0
We get rid of parentheses
-x^2-12x-12=0
We add all the numbers together, and all the variables
-1x^2-12x-12=0
a = -1; b = -12; c = -12;
Δ = b2-4ac
Δ = -122-4·(-1)·(-12)
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{-(-12)-4\sqrt{6}}{2*-1}=\frac{12-4\sqrt{6}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*-1}=\frac{12+4\sqrt{6}}{-2} $
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