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2x^2=324
We move all terms to the left:
2x^2-(324)=0
a = 2; b = 0; c = -324;
Δ = b2-4ac
Δ = 02-4·2·(-324)
Δ = 2592
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*2}=\frac{0-36\sqrt{2}}{4} =-\frac{36\sqrt{2}}{4} =-9\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*2}=\frac{0+36\sqrt{2}}{4} =\frac{36\sqrt{2}}{4} =9\sqrt{2} $
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