40=c^2-3c

Solution for 40=c^2-3c equation:

40=c^2-3c

We move all terms to the left:

40-(c^2-3c)=0

We get rid of parentheses

-c^2+3c+40=0

We add all the numbers together, and all the variables

-1c^2+3c+40=0

a = -1; b = 3; c = +40;
Δ = b2-4ac
Δ = 32-4·(-1)·40
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*-1}=\frac{-16}{-2}
=+8
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*-1}=\frac{10}{-2}
=-5
$