3x^2+25x+18=0

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Solution for 3x^2+25x+18=0 equation:



3x^2+25x+18=0
a = 3; b = 25; c = +18;
Δ = b2-4ac
Δ = 252-4·3·18
Δ = 409
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{409}}{2*3}=\frac{-25-\sqrt{409}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{409}}{2*3}=\frac{-25+\sqrt{409}}{6} $

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