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x^2+3x+9=27
We move all terms to the left:
x^2+3x+9-(27)=0
We add all the numbers together, and all the variables
x^2+3x-18=0
a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2} =3 $
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