x^2-8x+16=10-x

Solution for x^2-8x+16=10-x equation:

x^2-8x+16=10-x

We move all terms to the left:

x^2-8x+16-(10-x)=0

We add all the numbers together, and all the variables

x^2-8x-(-1x+10)+16=0

We get rid of parentheses

x^2-8x+1x-10+16=0

We add all the numbers together, and all the variables

x^2-7x+6=0

a = 1; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·1·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5}{2*1}=\frac{2}{2}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5}{2*1}=\frac{12}{2}
=6
$