3x^2 + 12x + 39 = 0

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You need to solve for x the quadratic equation, hence, you should use quadratic formula, such that:

`x_(1,2) = (-b+-sqrt(b^2 - 4ac))/(2a)`

Identifying coeficients a,b,c, yields:

`a = 3, b =12, c = 39`

`x_(1,2) = (-12+-sqrt(12^2 - 4*3*39))/(2*3)`

`x_(1,2) = (-12+-sqrt(144 - 468))/6`

`x_(1,2) = (-12+-sqrt(-324))/6`

Since the problem does not specify the nature of the solutions to equation, hence, you may also consider complex roots, such that:

`x_(1,2) = (-12+-18*i)/6`

Notice that complex number theory states that `sqrt(-1) = i.`

`x_(1,2) = (-2+-3*i)`

**Hence, evaluating the complec solutions to the given quadratic equation, yields `x_(1,2) = (-2+-3*i)` .**

`3x^2+12x+39=0`

`x^2+4x+13=0`

`x^2+4x +4+9=0`

`x^2+4x+4=-9`

`(x+2)^2=-9`

`x+2=+-3i`

`x=-2+-3i`

indeed `Delta= 16- 4 xx 13= 16-52=-36 <0`

So has two complex coniugate roots

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