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r^2=27
We move all terms to the left:
r^2-(27)=0
a = 1; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·1·(-27)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{3}}{2*1}=\frac{0-6\sqrt{3}}{2} =-\frac{6\sqrt{3}}{2} =-3\sqrt{3} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*1}=\frac{0+6\sqrt{3}}{2} =\frac{6\sqrt{3}}{2} =3\sqrt{3} $
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