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3t^2+8t-35=0
a = 3; b = 8; c = -35;
Δ = b2-4ac
Δ = 82-4·3·(-35)
Δ = 484
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{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*3}=\frac{-30}{6} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*3}=\frac{14}{6} =2+1/3 $
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