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15-6x^2=0
a = -6; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-6)·15
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{10}}{2*-6}=\frac{0-6\sqrt{10}}{-12} =-\frac{6\sqrt{10}}{-12} =-\frac{\sqrt{10}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{10}}{2*-6}=\frac{0+6\sqrt{10}}{-12} =\frac{6\sqrt{10}}{-12} =\frac{\sqrt{10}}{-2} $
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