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2b^2=27
We move all terms to the left:
2b^2-(27)=0
a = 2; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·2·(-27)
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{6}}{2*2}=\frac{0-6\sqrt{6}}{4} =-\frac{6\sqrt{6}}{4} =-\frac{3\sqrt{6}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{6}}{2*2}=\frac{0+6\sqrt{6}}{4} =\frac{6\sqrt{6}}{4} =\frac{3\sqrt{6}}{2} $
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