2j2+7j+6=0

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Solution for 2j2+7j+6=0 equation: 



2j^2+7j+6=0

a = 2; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·2·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*2}=\frac{-8}{4}
=-2
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*2}=\frac{-6}{4}
=-1+1/2
$

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