10x+4.9x^2=0

Solution for 10x+4.9x^2=0 equation:

10x+4.9x^2=0

a = 4.9; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·4.9·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*4.9}=\frac{-20}{9.8}
=-2+0.4/9.8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*4.9}=\frac{0}{9.8}
=0
$