# How do I use the quadratic formula to solve #3x^2 - 6 = 4x#?

##### 1 Answer

#### Explanation:

Given:

#3x^2-6 = 4x#

Subtract

#3x^2-4x-6 = 0#

This is now in standard form:

#ax^2+bx+c = 0#

with

So we can apply the quadratic formula to find:

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

#color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(3))(color(blue)(-6))))/(2(color(blue)(3)))#

#color(white)(x) = (4+-sqrt(16+72))/6#

#color(white)(x) = (4+-sqrt(88))/6#

#color(white)(x) = (4+-sqrt(2^2 * 22))/6#

#color(white)(x) = (4+-2sqrt(22))/6#

#color(white)(x) = (2+-sqrt(22))/3#

#color(white)(x) = 2/3+-sqrt(22)/3#

**Bonus**

Notice in the derivation above that we had a factor

This always happens when

In such cases we can instead think of the original quadratic as taking the form:

#ax^2+2dx+c = 0#

Then the solutions can be expressed as:

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

In our example

#x = (-(color(blue)(-2))+-sqrt((color(blue)(-2))^2-(color(blue)(3))(color(blue)(-6))))/(color(blue)(3))#

#color(white)(x) = (2+-sqrt(4+18))/3#

#color(white)(x) = (2+-sqrt(22))/3#

#color(white)(x) = 2/3+-sqrt(22)/3#

I don't know how widely this variant of the quadratic formula is used.

If

#x^2+2dx+c = 0#

are:

#x = -d+-sqrt(d^2-4c)#