0=21x+4.9x^2

Solution for 0=21x+4.9x^2 equation:

0=21x+4.9x^2

We move all terms to the left:

0-(21x+4.9x^2)=0

We add all the numbers together, and all the variables

-(21x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-21x=0

a = -4.9; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·(-4.9)·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*-4.9}=\frac{0}{-9.8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*-4.9}=\frac{42}{-9.8}
=-4+2.8/9.8
$