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17=35t+-t^2
We move all terms to the left:
17-(35t+-t^2)=0
We use the square of the difference formula
-(35t-t^2)+17=0
We get rid of parentheses
t^2-35t+17=0
a = 1; b = -35; c = +17;
Δ = b2-4ac
Δ = -352-4·1·17
Δ = 1157
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{-(-35)-\sqrt{1157}}{2*1}=\frac{35-\sqrt{1157}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1157}}{2*1}=\frac{35+\sqrt{1157}}{2} $
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