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9t^2+22t+9=0
a = 9; b = 22; c = +9;
Δ = b2-4ac
Δ = 222-4·9·9
Δ = 160
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-4\sqrt{10}}{2*9}=\frac{-22-4\sqrt{10}}{18} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+4\sqrt{10}}{2*9}=\frac{-22+4\sqrt{10}}{18} $
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