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16t^2+9t-82=0
a = 16; b = 9; c = -82;
Δ = b2-4ac
Δ = 92-4·16·(-82)
Δ = 5329
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{5329}=73$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-73}{2*16}=\frac{-82}{32} =-2+9/16 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+73}{2*16}=\frac{64}{32} =2 $
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