$x^2+29=10x+3$
take the 3 to the left-hand side
$ x^2+26=10x $
now take the 10x to the other side
$ x^2-10x+26=0 $
now to solve this you can pick many ways, though i will pick using the quadratic formula
$ _2x_1=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $
where $a=1$ ; $b=-10$ ; $c=26$
just plug and chug
$ _2x_1=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:26}}{2\cdot \:1} $
$ _2x_1=\frac{-\left(-10\right)\pm \iota \sqrt{10^2-104}} {2\cdot \:1} $
$ _2x_1=\frac{-\left(-10\right)\pm \iota \sqrt{4} } {2\cdot \:1} $
$ _2x_1=\frac{-\left(-10\right)\pm \:2\iota }{2\cdot \:1} $
that gives us $ x_1=\frac{10+2\iota }{2\cdot \:1} \ \ \ \Rightarrow \ \ \ \frac{10+2\iota }{2} \ \ \ \Rightarrow \ \ \ \frac{\cancel{2}\left(5+\iota \right)}
{\cancel{2}} $
finally, we get $x_1==5+\iota $
second root will be as following:
$x_2=\frac{10-2\iota }{2\cdot \:1} \ \ \ \Rightarrow \ \ \ \frac{10-2\iota a}{2} \ \ \ \Rightarrow \ \ \ \frac{\cancel{2}\left(5-\iota \right)}{\cancel{2}}$
thus the second root is $ x_2=5-\iota $