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3t^2+22t-16=0
a = 3; b = 22; c = -16;
Δ = b2-4ac
Δ = 222-4·3·(-16)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-26}{2*3}=\frac{-48}{6} =-8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+26}{2*3}=\frac{4}{6} =2/3 $
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