-w^2+11w-28=0

Solution for -w^2+11w-28=0 equation:

-w^2+11w-28=0

We add all the numbers together, and all the variables

-1w^2+11w-28=0

a = -1; b = 11; c = -28;
Δ = b2-4ac
Δ = 112-4·(-1)·(-28)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*-1}=\frac{-14}{-2}
=+7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*-1}=\frac{-8}{-2}
=+4
$