4.9t^2-19t=-13

Solution for 4.9t^2-19t=-13 equation:

4.9t^2-19t=-13

We move all terms to the left:

4.9t^2-19t-(-13)=0

We add all the numbers together, and all the variables

4.9t^2-19t+13=0

a = 4.9; b = -19; c = +13;
Δ = b2-4ac
Δ = -192-4·4.9·13
Δ = 106.2
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{-(-19)-\sqrt{106.2}}{2*4.9}=\frac{19-\sqrt{106.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{106.2}}{2*4.9}=\frac{19+\sqrt{106.2}}{9.8}
$