j2-17j+72=0

Solution for j2-17j+72=0 equation:

j2-17j+72=0

We add all the numbers together, and all the variables

j^2-17j+72=0

a = 1; b = -17; c = +72;
Δ = b2-4ac
Δ = -172-4·1·72
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-1}{2*1}=\frac{16}{2}
=8
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+1}{2*1}=\frac{18}{2}
=9
$