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6t^2-28t+22=0
a = 6; b = -28; c = +22;
Δ = b2-4ac
Δ = -282-4·6·22
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-16}{2*6}=\frac{12}{12} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+16}{2*6}=\frac{44}{12} =3+2/3 $
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