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

a = 4.9; b = -19; c = -13;
Δ = b2-4ac
Δ = -192-4·4.9·(-13)
Δ = 615.8
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{615.8}}{2*4.9}=\frac{19-\sqrt{615.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{615.8}}{2*4.9}=\frac{19+\sqrt{615.8}}{9.8}
$