-b^2-6b+40=0

Solution for -b^2-6b+40=0 equation:

-b^2-6b+40=0

We add all the numbers together, and all the variables

-1b^2-6b+40=0

a = -1; b = -6; c = +40;
Δ = b2-4ac
Δ = -62-4·(-1)·40
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*-1}=\frac{-8}{-2}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*-1}=\frac{20}{-2}
=-10
$