a^2-10a+8=0

Solution for a^2-10a+8=0 equation:

a^2-10a+8=0

a = 1; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·1·8
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{17}}{2*1}=\frac{10-2\sqrt{17}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{17}}{2*1}=\frac{10+2\sqrt{17}}{2}
$