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2b^2=225
We move all terms to the left:
2b^2-(225)=0
a = 2; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·2·(-225)
Δ = 1800
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{2}}{2*2}=\frac{0-30\sqrt{2}}{4} =-\frac{30\sqrt{2}}{4} =-\frac{15\sqrt{2}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{2}}{2*2}=\frac{0+30\sqrt{2}}{4} =\frac{30\sqrt{2}}{4} =\frac{15\sqrt{2}}{2} $
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