9t^2+4t-4.95=0

Solution for 9t^2+4t-4.95=0 equation:

9t^2+4t-4.95=0

a = 9; b = 4; c = -4.95;
Δ = b2-4ac
Δ = 42-4·9·(-4.95)
Δ = 194.2
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{-(4)-\sqrt{194.2}}{2*9}=\frac{-4-\sqrt{194.2}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{194.2}}{2*9}=\frac{-4+\sqrt{194.2}}{18}
$