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3t^2+29t+40=0
a = 3; b = 29; c = +40;
Δ = b2-4ac
Δ = 292-4·3·40
Δ = 361
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{361}=19$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*3}=\frac{-48}{6} =-8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*3}=\frac{-10}{6} =-1+2/3 $
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