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

Solution for -4.9x^2-2x+100=0 equation:

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

a = -4.9; b = -2; c = +100;
Δ = b2-4ac
Δ = -22-4·(-4.9)·100
Δ = 1964
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:

$\sqrt{\Delta}=\sqrt{1964}=\sqrt{4*491}=\sqrt{4}*\sqrt{491}=2\sqrt{491}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{491}}{2*-4.9}=\frac{2-2\sqrt{491}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{491}}{2*-4.9}=\frac{2+2\sqrt{491}}{-9.8}
$