9t2+17t-2=0

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Solution for 9t2+17t-2=0 equation:



9t^2+17t-2=0
a = 9; b = 17; c = -2;
Δ = b2-4ac
Δ = 172-4·9·(-2)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*9}=\frac{-36}{18} =-2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*9}=\frac{2}{18} =1/9 $

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