9t^2+t-6=0

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Solution for 9t^2+t-6=0 equation: 



9t^2+t-6=0

a = 9; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·9·(-6)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{217}}{2*9}=\frac{-1-\sqrt{217}}{18}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{217}}{2*9}=\frac{-1+\sqrt{217}}{18}
$

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