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3x^2=225
We move all terms to the left:
3x^2-(225)=0
a = 3; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·3·(-225)
Δ = 2700
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{3}}{2*3}=\frac{0-30\sqrt{3}}{6} =-\frac{30\sqrt{3}}{6} =-5\sqrt{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{3}}{2*3}=\frac{0+30\sqrt{3}}{6} =\frac{30\sqrt{3}}{6} =5\sqrt{3} $
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