8t^2+9t-9=0

Solution for 8t^2+9t-9=0 equation:

8t^2+9t-9=0

a = 8; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·8·(-9)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{41}}{2*8}=\frac{-9-3\sqrt{41}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{41}}{2*8}=\frac{-9+3\sqrt{41}}{16}
$