Solve by using the quadratic formula 3x^2+5=-2x

Hello from MrBillDoesMath!

  -1/3 + i* (1/3) * sqrt(14)  and  -1/3 - i* (1/3) * sqrt(14)

Discussion :

Recall the solutions of the quadratic equation  "ax^2 + bx + c = 0" are given by the formula

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

Rewriting the given equation in this for gives

3x^2 + 2x + 5 = 0

so a = 3, b = 2, and c = 5.  Substituting the values in the formula gives

x =  (- 2 + sqrt ( 2^2 - 4*3*5)  ) / 2a         #1

and

x =  (- 2 - sqrt ( 2^2 - 4*3*5)  ) / 2a       #2

Simplifying, #1 = ( -2   + sqrt(4-60) )/ (2 * 3)  =

                           -2/(2*3)  +  (sqrt(-56)) / (2*3) =

                            -1/3  + i * (sqrt (56)) /6 =

                            -1/3 +  i * (sqrt(4 *14)) /6 =

                            -1/3  + i * (sqrt(4)/6) * sqrt(14)  =

                            -1/3 + i * (2/6) * sqrt(14) =

                            -1/3 + i* (1/3) * sqrt(14)

Similarly, the solution of # 2=   -1/3 - i* (1/3) * sqrt(14). The only difference between the two solutions is the minus sign boldfaced above.

Regards, MrB

there are no figures how am i supposed to answer this question

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