-b^2+8b-13=0

Solution for -b^2+8b-13=0 equation:

-b^2+8b-13=0

We add all the numbers together, and all the variables

-1b^2+8b-13=0

a = -1; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·(-1)·(-13)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{3}}{2*-1}=\frac{-8-2\sqrt{3}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{3}}{2*-1}=\frac{-8+2\sqrt{3}}{-2}
$