Ax^2-4x+5=8

Solution for Ax^2-4x+5=8 equation:

A^2-4A+5=8

We move all terms to the left:

A^2-4A+5-(8)=0

We add all the numbers together, and all the variables

A^2-4A-3=0

a = 1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·1·(-3)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{7}}{2*1}=\frac{4-2\sqrt{7}}{2}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{7}}{2*1}=\frac{4+2\sqrt{7}}{2}
$