5j^2+36j+7=0

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Solution for 5j^2+36j+7=0 equation:



5j^2+36j+7=0
a = 5; b = 36; c = +7;
Δ = b2-4ac
Δ = 362-4·5·7
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-34}{2*5}=\frac{-70}{10} =-7 $
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+34}{2*5}=\frac{-2}{10} =-1/5 $

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