2x^2+25x+4.9=0

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Solution for 2x^2+25x+4.9=0 equation:



2x^2+25x+4.9=0
a = 2; b = 25; c = +4.9;
Δ = b2-4ac
Δ = 252-4·2·4.9
Δ = 585.8
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{-(25)-\sqrt{585.8}}{2*2}=\frac{-25-\sqrt{585.8}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{585.8}}{2*2}=\frac{-25+\sqrt{585.8}}{4} $

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