# Solve 3x^2+x+7=0

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`3x^2 + x + 7= 0` .

The roots are given by the following formula:

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

`==gt x1= (-1+sqrt(1-4*7*3))/(2*3) `

`==gt x2= (-1+sqrt(-83))/6 = (-1+sqrt83*i)/6 `

`==gt x2= (-1-sqrt83*i)/6`

`` Then we have two complex roots: `(-1+sqrt83*i)/6 and (-1-sqrt83*i)/6.`

There are two best approaches you can take to obtain your solution. First, you could enter the function into a graphing calculator and have it solve for the roots---but most teachers are going to want you to do it by hand!

That being said, the quadratic equation will be your best bet. Remember the equation states for any quadratic ax^2 + bx +c,

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

(where sqrt(x) is a shorthand for square root, +/- is plus or minus, and the carat (^) denotes an exponent)

Looking at our equation, we obtain that a= 3, b =1, and c = 7. All we need to do is substitute in these values to our equation:

x = (-1 +/- sqrt[1^2-4(3)(7)]) / 2(3)

x= (-1 +/- sqrt[1-84]) / 6

x= (-1 +/- sqrt[-83]) / 6 --> x = (-1 +/- i*sqrt[83])/6.

So our solutions are x = (-1 + i*sqrt[83])/6 and (-1 - sqrt[83])/6.