0=4.9x^2+40x-50

Solution for 0=4.9x^2+40x-50 equation:

0=4.9x^2+40x-50

We move all terms to the left:

0-(4.9x^2+40x-50)=0

We add all the numbers together, and all the variables

-(4.9x^2+40x-50)=0

We get rid of parentheses

-4.9x^2-40x+50=0

a = -4.9; b = -40; c = +50;
Δ = b2-4ac
Δ = -402-4·(-4.9)·50
Δ = 2580
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2580}=\sqrt{4*645}=\sqrt{4}*\sqrt{645}=2\sqrt{645}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{645}}{2*-4.9}=\frac{40-2\sqrt{645}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{645}}{2*-4.9}=\frac{40+2\sqrt{645}}{-9.8}
$