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16t^2+75t+2=0
a = 16; b = 75; c = +2;
Δ = b2-4ac
Δ = 752-4·16·2
Δ = 5497
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{-(75)-\sqrt{5497}}{2*16}=\frac{-75-\sqrt{5497}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+\sqrt{5497}}{2*16}=\frac{-75+\sqrt{5497}}{32} $
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