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15=16t^2
We move all terms to the left:
15-(16t^2)=0
a = -16; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-16)·15
Δ = 960
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{15}}{2*-16}=\frac{0-8\sqrt{15}}{-32} =-\frac{8\sqrt{15}}{-32} =-\frac{\sqrt{15}}{-4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{15}}{2*-16}=\frac{0+8\sqrt{15}}{-32} =\frac{8\sqrt{15}}{-32} =\frac{\sqrt{15}}{-4} $
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