j^2-16j+39=0

Solution for j^2-16j+39=0 equation:

j^2-16j+39=0

a = 1; b = -16; c = +39;
Δ = b2-4ac
Δ = -162-4·1·39
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-10}{2*1}=\frac{6}{2}
=3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+10}{2*1}=\frac{26}{2}
=13
$