-4.9x^2+400=0

Solution for -4.9x^2+400=0 equation:

-4.9x^2+400=0

a = -4.9; b = 0; c = +400;
Δ = b2-4ac
Δ = 02-4·(-4.9)·400
Δ = 7840
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:

$\sqrt{\Delta}=\sqrt{7840}=\sqrt{16*490}=\sqrt{16}*\sqrt{490}=4\sqrt{490}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{490}}{2*-4.9}=\frac{0-4\sqrt{490}}{-9.8}
=-\frac{4\sqrt{490}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{490}}{2*-4.9}=\frac{0+4\sqrt{490}}{-9.8}
=\frac{4\sqrt{490}}{-9.8}
$