Algebra

$x^2+29=10x+3$

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 $