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-16t^2+4=0
a = -16; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-16)·4
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-16}=\frac{-16}{-32} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-16}=\frac{16}{-32} =-1/2 $
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