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-9t^2+10t-1=0
a = -9; b = 10; c = -1;
Δ = b2-4ac
Δ = 102-4·(-9)·(-1)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*-9}=\frac{-18}{-18} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*-9}=\frac{-2}{-18} =1/9 $
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