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15=3x^2
We move all terms to the left:
15-(3x^2)=0
a = -3; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-3)·15
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{5}}{2*-3}=\frac{0-6\sqrt{5}}{-6} =-\frac{6\sqrt{5}}{-6} =-\frac{\sqrt{5}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{5}}{2*-3}=\frac{0+6\sqrt{5}}{-6} =\frac{6\sqrt{5}}{-6} =\frac{\sqrt{5}}{-1} $
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