Y=x^2+10x+13

Solution for Y=x^2+10x+13 equation:

=Y^2+10Y+13

We move all terms to the left:

-(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}
$