Question

Solve. (m^2+m) (m^2+m-3)=28

Answer 1

Please see the attachment

Answer 2

Answer:

$(m^2+m)(m^2+m-3)=28 \\x^4+2x^3-2x^2-3x=28\\ x^4+2x^3-2x^2-3x-28=0 \\(x^2+x-7)(x^2+4x+4)=0\\x=\frac{-1-\sqrt{29} }{2}, \frac{-1+\sqrt{29} }{2}$

Since factor $(x^2+4x+4)$ has a discriminant less than 0, it has no real solutions. Therefore, there are only two real solutions, and can only be obtain through quadratic formula as the discriminant for $(x^2+x-7)$ is not a perfect square, indicating that solutions are not rational.

Hope it helps :)

Step-by-step explanation: