-b^2+8b-6=0

Solution for -b^2+8b-6=0 equation:

-b^2+8b-6=0

We add all the numbers together, and all the variables

-1b^2+8b-6=0

a = -1; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·(-1)·(-6)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*-1}=\frac{-8-2\sqrt{10}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*-1}=\frac{-8+2\sqrt{10}}{-2}
$