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14t^2+58t+8=0
a = 14; b = 58; c = +8;
Δ = b2-4ac
Δ = 582-4·14·8
Δ = 2916
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{2916}=54$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-54}{2*14}=\frac{-112}{28} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+54}{2*14}=\frac{-4}{28} =-1/7 $
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