Question

From the previous step we have erx(6r2 + 3r − 3) = 0. Since erx is never 0, then we know 6r2 + 3r − 3 = 0. Solving the above equation gives the solutions to the quadratic equation.

Answer 1

Answer:

6r2+3r-3=0

two solutions are possible

r= -1

r=1/2

Answer 2

Answer:

 r = - 1      or   r = 1/2

Step-by-step explanation:

From the quadratic equation;

                $6r^{2} + 3r - 3 = 0$

          $6r^{2}$ + 6r - 3r - 3 = 0

           ( $6r^{2}$ + 6r) - (3r + 3) = 0

         6r(r + 1)  - 3 (r + 1) = 0

         ⇒ (r + 1) or (6r - 3) = 0

         r + 1 = 0     or 6r - 3 = 0

         r = - 1      or      6r = 3

         r = - 1      or   r = 1/2

Thus, the solutions to the equation are; r = - 1 or r = 1/2 since erx is never 0.