0=4.9x^2-50x+40

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

0=4.9x^2-50x+40

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -4.9; b = 50; c = -40;
Δ = b2-4ac
Δ = 502-4·(-4.9)·(-40)
Δ = 1716
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:

$\sqrt{\Delta}=\sqrt{1716}=\sqrt{4*429}=\sqrt{4}*\sqrt{429}=2\sqrt{429}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{429}}{2*-4.9}=\frac{-50-2\sqrt{429}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{429}}{2*-4.9}=\frac{-50+2\sqrt{429}}{-9.8}
$