Question

What is x if `(4x+3)^2=7`?

Answer 1

`(4x+3)^2=7`

`16x^2+24x+9-7=0`

`16x^2+24x+2=0`

`8x^2+12x+1=0`

`=((-12)+-sqrt((12^2)-4(8*1)))/(2*8)`

As the quadratic formula says, from here on I leave you

Answer 2

`x=(-3+-sqrt7)/"4"`

Explanation:

First we expand the left hand side of the equation, by multiplying
`(4x+3)(4x+3)=7`

Giving us
`16x^2+12x+12x+9=7`

Now we get all terms to one side and combine
`16x^2+24x+2=0`

Now we can divide all by the constant `2`, giving us
`2(8x^2+12x+1)=0`

Now we can divide `2` on both sides, should look like this
`2/2(8x^2+12x+1)=0/2`

Which simplifies to
`8x^2+12x+1=0`

Now this does not factor, we have to use the Quadratic Formula which is,
Quadratic Formula is
`(-b +- sqrt(b^2-4ac))/"2a"`

Where
`a=8`
`b=12`
`c=1`

Now we just plug it in
`(-(12)+-sqrt(12^2-4(8)(1)))/"2(8)"`

You can split this up like so
`-(12)/"2(8)"+-(sqrt(12^2-4(8)(1)))/"2(8)"`

After multipying and combining like terms we get
`-(12)/16+-sqrt(144-32)/"16"`

This then becomes
`-3/4+-sqrt(112)/16`

`-3/4+-(sqrt(16)*sqrt(7))/16`

`-3/4+-(4*sqrt(7))/16`

`-3/4+-sqrt7/4`

They have common denominators so we can have this
`x=(-3+-sqrt7)/4`

Answer 3

`x = frac(- 3 pm sqrt(7))(4)`

Explanation:

We have: `(4x + 3)^(2) = 7`

This problem can be done without the use of the quadratic formula.

Let's take the square root of both sides of the equation:

`Rightarrow 4x + 3 = pm sqrt(7)`

Subtracting `3` from both sides:

`Rightarrow 4x = - 3 pm sqrt(7)`

Dividing both sides by `4`:

`therefore x = frac(- 3 pm sqrt(7))(4)`