9t^2-22+9=0

Solution for 9t^2-22+9=0 equation:

9t^2-22+9=0

We add all the numbers together, and all the variables

9t^2-13=0

a = 9; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·9·(-13)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{13}}{2*9}=\frac{0-6\sqrt{13}}{18}
=-\frac{6\sqrt{13}}{18}
=-\frac{\sqrt{13}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{13}}{2*9}=\frac{0+6\sqrt{13}}{18}
=\frac{6\sqrt{13}}{18}
=\frac{\sqrt{13}}{3}
$