4.9t^2-16t-15=0

Solution for 4.9t^2-16t-15=0 equation:

4.9t^2-16t-15=0

a = 4.9; b = -16; c = -15;
Δ = b2-4ac
Δ = -162-4·4.9·(-15)
Δ = 550
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:

$\sqrt{\Delta}=\sqrt{550}=\sqrt{25*22}=\sqrt{25}*\sqrt{22}=5\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-5\sqrt{22}}{2*4.9}=\frac{16-5\sqrt{22}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+5\sqrt{22}}{2*4.9}=\frac{16+5\sqrt{22}}{9.8}
$