0=4.9x^2+49x

Solution for 0=4.9x^2+49x equation:

0=4.9x^2+49x

We move all terms to the left:

0-(4.9x^2+49x)=0

We add all the numbers together, and all the variables

-(4.9x^2+49x)=0

We get rid of parentheses

-4.9x^2-49x=0

a = -4.9; b = -49; c = 0;
Δ = b2-4ac
Δ = -492-4·(-4.9)·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-49}{2*-4.9}=\frac{0}{-9.8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+49}{2*-4.9}=\frac{98}{-9.8}
=-10
$