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6t^2-22t+12=0
a = 6; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·6·12
Δ = 196
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{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*6}=\frac{8}{12} =2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*6}=\frac{36}{12} =3 $
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