4.9x^2+17x=0

Solution for 4.9x^2+17x=0 equation:

4.9x^2+17x=0

a = 4.9; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·4.9·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*4.9}=\frac{-34}{9.8}
=-3+2.875/6.125
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*4.9}=\frac{0}{9.8}
=0
$