10-4a+a^2=8

Solution for 10-4a+a^2=8 equation:

10-4a+a^2=8

We move all terms to the left:

10-4a+a^2-(8)=0

We add all the numbers together, and all the variables

a^2-4a+2=0

a = 1; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·1·2
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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