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-8t^2+16t=0
a = -8; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-8)·0
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-8}=\frac{-32}{-16} =+2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-8}=\frac{0}{-16} =0 $
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