7j2+12j+5=0

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Solution for 7j2+12j+5=0 equation: 



7j^2+12j+5=0

a = 7; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·7·5
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*7}=\frac{-14}{14}
=-1
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*7}=\frac{-10}{14}
=-5/7
$

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