-b^2+28b-132=0

Solution for -b^2+28b-132=0 equation:

-b^2+28b-132=0

We add all the numbers together, and all the variables

-1b^2+28b-132=0

a = -1; b = 28; c = -132;
Δ = b2-4ac
Δ = 282-4·(-1)·(-132)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-16}{2*-1}=\frac{-44}{-2}
=+22
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+16}{2*-1}=\frac{-12}{-2}
=+6
$