-b^2-5b+100=0

Solution for -b^2-5b+100=0 equation:

-b^2-5b+100=0

We add all the numbers together, and all the variables

-1b^2-5b+100=0

a = -1; b = -5; c = +100;
Δ = b2-4ac
Δ = -52-4·(-1)·100
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{17}}{2*-1}=\frac{5-5\sqrt{17}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{17}}{2*-1}=\frac{5+5\sqrt{17}}{-2}
$