# How do you solve #x^2+x=20#?

##### 2 Answers

#### Explanation:

Here are a few methods:

**Method 1 - Fishing for factors**

Note that

#x(x+1) = 20#

So we want to find a number

It should not take too much guessing to find:

#color(blue)(4)(color(blue)(4)+1) = (4)(5) = 20#

So

This is a quadratic equation, so should normally have two solutions, so what is the other?

With a bit of thought, we find:

#((color(blue)(-5))((color(blue)(-5))+1) = (-5)(-4) = 20#

So

**Method 2 - Completing the square**

Given:

#x^2+x=20#

Note that in general:

#(x+p)^2 = x^2+2px+p^2#

So to make

#x^2+x+1/4 = 81/4#

Both sides are now perfect squares:

#(x+1/2)^2 = (9/2)^2#

Hence:

#x+1/2 = +-9/2#

Subtracting

#x = -1/2+-9/2#

That is:

#x = 4" "# or#" "x = -5#

**Method 3 - Quadratic formula**

Given:

#x^2+x=20#

Subtract

#x^2+x-20 = 0#

This is in the standard form:

#ax^2+bx+c = 0#

with

It has solutions given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-1+-sqrt((-1)^2-4(1)(-20)))/(2*1)#

#color(white)(x) = (-1+-sqrt(1+80))/2#

#color(white)(x) = (-1+-9)/2#

That is:

#x = 4" "# or#" "x = -5#

#### Explanation:

Method 4. Solving by new Transforming Method (Google Search)

Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (- b = - 1) and the product (c = - 20).

They are: