0=x^2-12x+26

Solution for 0=x^2-12x+26 equation:

0=x^2-12x+26

We move all terms to the left:

0-(x^2-12x+26)=0

We add all the numbers together, and all the variables

-(x^2-12x+26)=0

We get rid of parentheses

-x^2+12x-26=0

We add all the numbers together, and all the variables

-1x^2+12x-26=0

a = -1; b = 12; c = -26;
Δ = b2-4ac
Δ = 122-4·(-1)·(-26)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{10}}{2*-1}=\frac{-12-2\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{10}}{2*-1}=\frac{-12+2\sqrt{10}}{-2}
$