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1-16t^2=0
a = -16; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-16)·1
Δ = 64
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-16}=\frac{-8}{-32} =1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-16}=\frac{8}{-32} =-1/4 $
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