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5t^2+9t-10=0
a = 5; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·5·(-10)
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{-(9)-\sqrt{281}}{2*5}=\frac{-9-\sqrt{281}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{281}}{2*5}=\frac{-9+\sqrt{281}}{10} $
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