j^2-15j+36=0

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Solution for j^2-15j+36=0 equation: 



j^2-15j+36=0

a = 1; b = -15; c = +36;
Δ = b2-4ac
Δ = -152-4·1·36
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{-(-15)-9}{2*1}=\frac{6}{2}
=3
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*1}=\frac{24}{2}
=12
$

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