20=-4.9x^2-42x+100

Solution for 20=-4.9x^2-42x+100 equation:

20=-4.9x^2-42x+100

We move all terms to the left:

20-(-4.9x^2-42x+100)=0

We get rid of parentheses

4.9x^2+42x-100+20=0

We add all the numbers together, and all the variables

4.9x^2+42x-80=0

a = 4.9; b = 42; c = -80;
Δ = b2-4ac
Δ = 422-4·4.9·(-80)
Δ = 3332
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:

$\sqrt{\Delta}=\sqrt{3332}=\sqrt{196*17}=\sqrt{196}*\sqrt{17}=14\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-14\sqrt{17}}{2*4.9}=\frac{-42-14\sqrt{17}}{9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+14\sqrt{17}}{2*4.9}=\frac{-42+14\sqrt{17}}{9.8}
$