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16t^2-37t+17=0
a = 16; b = -37; c = +17;
Δ = b2-4ac
Δ = -372-4·16·17
Δ = 281
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-37)-\sqrt{281}}{2*16}=\frac{37-\sqrt{281}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-37)+\sqrt{281}}{2*16}=\frac{37+\sqrt{281}}{32} $
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