j2-8j+10=0

Solution for j2-8j+10=0 equation:

j2-8j+10=0

We add all the numbers together, and all the variables

j^2-8j+10=0

a = 1; b = -8; c = +10;
Δ = b2-4ac
Δ = -82-4·1·10
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{6}}{2*1}=\frac{8-2\sqrt{6}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{6}}{2*1}=\frac{8+2\sqrt{6}}{2}
$