4.905x^2+10x-5.1=0

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Solution for 4.905x^2+10x-5.1=0 equation:



4.905x^2+10x-5.1=0
a = 4.905; b = 10; c = -5.1;
Δ = b2-4ac
Δ = 102-4·4.905·(-5.1)
Δ = 200.062
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{-(10)-\sqrt{200.062}}{2*4.905}=\frac{-10-\sqrt{200.062}}{9.81} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{200.062}}{2*4.905}=\frac{-10+\sqrt{200.062}}{9.81} $

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