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16t^2+63t+4=0
a = 16; b = 63; c = +4;
Δ = b2-4ac
Δ = 632-4·16·4
Δ = 3713
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{-(63)-\sqrt{3713}}{2*16}=\frac{-63-\sqrt{3713}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+\sqrt{3713}}{2*16}=\frac{-63+\sqrt{3713}}{32} $
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