4.9t^2+7t-21=0

Solution for 4.9t^2+7t-21=0 equation:

4.9t^2+7t-21=0

a = 4.9; b = 7; c = -21;
Δ = b2-4ac
Δ = 72-4·4.9·(-21)
Δ = 460.6
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{-(7)-\sqrt{460.6}}{2*4.9}=\frac{-7-\sqrt{460.6}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{460.6}}{2*4.9}=\frac{-7+\sqrt{460.6}}{9.8}
$