3x^2+23x+41=0

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Solution for 3x^2+23x+41=0 equation: 



3x^2+23x+41=0

a = 3; b = 23; c = +41;
Δ = b2-4ac
Δ = 232-4·3·41
Δ = 37
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{-(23)-\sqrt{37}}{2*3}=\frac{-23-\sqrt{37}}{6}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{37}}{2*3}=\frac{-23+\sqrt{37}}{6}
$

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