# What is the solution of 14x^2 + 3x - 7 = 0

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### 2 Answers

The solution of the equation 14x^2 + 3x - 7 = 0 has to be determined.

The roots of a quadratic equation ax^2 + bx + c = 0 are `(-b+-sqrt(b^2 - 4ac))/(2a)`

For the equation 14x^2 + 3x - 7 = 0, the roots are:

`(-3 +- sqrt(9 + 392))/28`

= `(-3 +- sqrt 401)/28`

**The roots of the equation `14x^2 + 3x - 7 = 0` are **`(-3 +- sqrt 401)/28`

The equation `14x^2+3x-7=0` can be written as

`x^2+3/14x-7/14=0`

or, `x^2+3/14x-1/2=0.`

Let roots of the above equation be a and b. So the above equation can be expressed as

`x^2+3/14x-1/2=(x-a)(x-b)`

or, `x^2+3/14x-1/2=x^2-(a+b)x+ab`

Comparing both sides we get

`(a+b)=-3/14` .........(1)

and `ab=-1/2` .........(2)

We know that `(a-b)^2=(a+b)^2-4ab`

Using (1) and (2) we get

`(a-b)^2=(-3/14)^2-4(-1/2)`

or, `(a-b)^2=9/196+2`

or, `(a-b)^2=401/196`

or, `(a-b)=+-sqrt(401)/14` .....(3)

If We take positive sign

Using (1) and (3) we get

`a=(-3+sqrt(401))/28`

and `b=(-3-sqrt(401))/28` .

If we take negative sign we get

`a=(-3-sqrt(401))/28`

and `b=(-3+sqrt(401))/28` .