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9t^2-18t+8=0
a = 9; b = -18; c = +8;
Δ = b2-4ac
Δ = -182-4·9·8
Δ = 36
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{36}=6$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*9}=\frac{12}{18} =2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*9}=\frac{24}{18} =1+1/3 $
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