j^2+17j=0

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Solution for j^2+17j=0 equation: 



j^2+17j=0

a = 1; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·1·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*1}=\frac{-34}{2}
=-17
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*1}=\frac{0}{2}
=0
$

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