j^2-4j-12=0

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Solution for j^2-4j-12=0 equation: 



j^2-4j-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


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


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$

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