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-16t^2+105t=34
We move all terms to the left:
-16t^2+105t-(34)=0
a = -16; b = 105; c = -34;
Δ = b2-4ac
Δ = 1052-4·(-16)·(-34)
Δ = 8849
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{-(105)-\sqrt{8849}}{2*-16}=\frac{-105-\sqrt{8849}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(105)+\sqrt{8849}}{2*-16}=\frac{-105+\sqrt{8849}}{-32} $
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