-x^2+22x=40

Solution for -x^2+22x=40 equation:

-x^2+22x=40

We move all terms to the left:

-x^2+22x-(40)=0

We add all the numbers together, and all the variables

-1x^2+22x-40=0

a = -1; b = 22; c = -40;
Δ = b2-4ac
Δ = 222-4·(-1)·(-40)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*-1}=\frac{-40}{-2}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*-1}=\frac{-4}{-2}
=+2
$