solve the equation: x^2 = 3x+40

We'll subtract 3x + 40 both
sides:

x^2 - 3x-40 = 0

We'll
apply quadratic formula:

x1 =
[-b+sqrt(b^2-4ac)]/2a

x2 =
[-b-sqrt(b^2-4ac)]/2a

We'll identify
a,b,c:

a=1

b=-3

c=-40

We'll
calculate delta:

delta =
b^2-4ac

delta = 9+160

delta =
169

Since delta is positive, we'll have 2 distinct real
roots of the equation x^2 - 3x-40 =
0.

x1=[3+sqrt(9+160)]/2

x1 =
(3+13)/2

x1 = 8

x2
=  -5

The real roots are: {-5 ;
8}.