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6d^2-15=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:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$d_{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} $$d_{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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