4.9t^2+20t-292=0

Solution for 4.9t^2+20t-292=0 equation:

4.9t^2+20t-292=0

a = 4.9; b = 20; c = -292;
Δ = b2-4ac
Δ = 202-4·4.9·(-292)
Δ = 6123.2
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{-(20)-\sqrt{6123.2}}{2*4.9}=\frac{-20-\sqrt{6123.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{6123.2}}{2*4.9}=\frac{-20+\sqrt{6123.2}}{9.8}
$