9t^2+92t-20=0

Solution for 9t^2+92t-20=0 equation:

9t^2+92t-20=0

a = 9; b = 92; c = -20;
Δ = b2-4ac
Δ = 922-4·9·(-20)
Δ = 9184
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:

$\sqrt{\Delta}=\sqrt{9184}=\sqrt{16*574}=\sqrt{16}*\sqrt{574}=4\sqrt{574}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{574}}{2*9}=\frac{-92-4\sqrt{574}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{574}}{2*9}=\frac{-92+4\sqrt{574}}{18}
$