19=4.9x^2+20x+4

Solution for 19=4.9x^2+20x+4 equation:

19=4.9x^2+20x+4

We move all terms to the left:

19-(4.9x^2+20x+4)=0

We get rid of parentheses

-4.9x^2-20x-4+19=0

We add all the numbers together, and all the variables

-4.9x^2-20x+15=0

a = -4.9; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·(-4.9)·15
Δ = 694
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-\sqrt{694}}{2*-4.9}=\frac{20-\sqrt{694}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{694}}{2*-4.9}=\frac{20+\sqrt{694}}{-9.8}
$