4.905t^2+5t-15=0

Solution for 4.905t^2+5t-15=0 equation:

4.905t^2+5t-15=0

a = 4.905; b = 5; c = -15;
Δ = b2-4ac
Δ = 52-4·4.905·(-15)
Δ = 319.3
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{-(5)-\sqrt{319.3}}{2*4.905}=\frac{-5-\sqrt{319.3}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{319.3}}{2*4.905}=\frac{-5+\sqrt{319.3}}{9.81}
$