2x^2+17x+6=0

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Solution for 2x^2+17x+6=0 equation:



2x^2+17x+6=0
a = 2; b = 17; c = +6;
Δ = b2-4ac
Δ = 172-4·2·6
Δ = 241
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{-(17)-\sqrt{241}}{2*2}=\frac{-17-\sqrt{241}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{241}}{2*2}=\frac{-17+\sqrt{241}}{4} $

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