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16t^2+194t+69=0
a = 16; b = 194; c = +69;
Δ = b2-4ac
Δ = 1942-4·16·69
Δ = 33220
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{33220}=\sqrt{4*8305}=\sqrt{4}*\sqrt{8305}=2\sqrt{8305}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(194)-2\sqrt{8305}}{2*16}=\frac{-194-2\sqrt{8305}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(194)+2\sqrt{8305}}{2*16}=\frac{-194+2\sqrt{8305}}{32} $
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