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6t^2+2t-28=0
a = 6; b = 2; c = -28;
Δ = b2-4ac
Δ = 22-4·6·(-28)
Δ = 676
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{676}=26$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-26}{2*6}=\frac{-28}{12} =-2+1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+26}{2*6}=\frac{24}{12} =2 $
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