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16t^2+5t=30
We move all terms to the left:
16t^2+5t-(30)=0
a = 16; b = 5; c = -30;
Δ = b2-4ac
Δ = 52-4·16·(-30)
Δ = 1945
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{-(5)-\sqrt{1945}}{2*16}=\frac{-5-\sqrt{1945}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1945}}{2*16}=\frac{-5+\sqrt{1945}}{32} $
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