2j^2+3j-9=0

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Solution for 2j^2+3j-9=0 equation: 



2j^2+3j-9=0

a = 2; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·2·(-9)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*2}=\frac{-12}{4}
=-3
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*2}=\frac{6}{4}
=1+1/2
$

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