4.9t^2-17t-10=0

Solution for 4.9t^2-17t-10=0 equation:

4.9t^2-17t-10=0

a = 4.9; b = -17; c = -10;
Δ = b2-4ac
Δ = -172-4·4.9·(-10)
Δ = 485
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{-(-17)-\sqrt{485}}{2*4.9}=\frac{17-\sqrt{485}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{485}}{2*4.9}=\frac{17+\sqrt{485}}{9.8}
$