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9t^2+44t-5=0
a = 9; b = 44; c = -5;
Δ = b2-4ac
Δ = 442-4·9·(-5)
Δ = 2116
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{2116}=46$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-46}{2*9}=\frac{-90}{18} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+46}{2*9}=\frac{2}{18} =1/9 $
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