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16t^2-38t+15=0
a = 16; b = -38; c = +15;
Δ = b2-4ac
Δ = -382-4·16·15
Δ = 484
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{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-22}{2*16}=\frac{16}{32} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+22}{2*16}=\frac{60}{32} =1+7/8 $
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