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9t^2-18t+2=0
a = 9; b = -18; c = +2;
Δ = b2-4ac
Δ = -182-4·9·2
Δ = 252
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{7}}{2*9}=\frac{18-6\sqrt{7}}{18} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{7}}{2*9}=\frac{18+6\sqrt{7}}{18} $
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