0=y^2-10y+13

Solution for 0=y^2-10y+13 equation:

0=y^2-10y+13

We move all terms to the left:

0-(y^2-10y+13)=0

We add all the numbers together, and all the variables

-(y^2-10y+13)=0

We get rid of parentheses

-y^2+10y-13=0

We add all the numbers together, and all the variables

-1y^2+10y-13=0

a = -1; b = 10; c = -13;
Δ = b2-4ac
Δ = 102-4·(-1)·(-13)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{3}}{2*-1}=\frac{-10-4\sqrt{3}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{3}}{2*-1}=\frac{-10+4\sqrt{3}}{-2}
$