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6t^2+31t+5=0
a = 6; b = 31; c = +5;
Δ = b2-4ac
Δ = 312-4·6·5
Δ = 841
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{841}=29$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-29}{2*6}=\frac{-60}{12} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+29}{2*6}=\frac{-2}{12} =-1/6 $
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