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16t^2-24t+8=0
a = 16; b = -24; c = +8;
Δ = b2-4ac
Δ = -242-4·16·8
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8}{2*16}=\frac{16}{32} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*16}=\frac{32}{32} =1 $
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