4.9x^2+1.8+20x=0

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Solution for 4.9x^2+1.8+20x=0 equation:



4.9x^2+1.8+20x=0
a = 4.9; b = 20; c = +1.8;
Δ = b2-4ac
Δ = 202-4·4.9·1.8
Δ = 364.72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-\sqrt{364.72}}{2*4.9}=\frac{-20-\sqrt{364.72}}{9.8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{364.72}}{2*4.9}=\frac{-20+\sqrt{364.72}}{9.8} $

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