Answer:


Let (x^{2} +
x) = m




∴ m(m – 2) = 24


∴ m^{2} – 2m = 24


∴ m^{2}  2m  24 = 0


∴ m^{2} – 6m + 4m – 24 = 0


∴ m(m – 6) + 4( m – 6 ) = 0


∴ (m – 6) (m + 4 ) = 0


∴ m – 6 = 0

∴ m + 4 = 0

∴ m = 6

∴ m =  4

Re substituting m = (x^{2}
+ x)




∴ x^{2} + x = 6

∴ x^{2} + x =  4



∴ x^{2} + x – 6 = 0

∴ x^{ 2 }+ x + 4 = 0



∴ x^{2} + 3x – 2x – 6 = 0

Here, a = 1 , b = 1, c = 4



∴ x(x + 3) – 2(x + 3) = 0

∴ b^{2} – 4ac = 1^{2} – 4 (1) (4)



∴ (x + 3 ) ( x – 2 ) = 0

∴ b^{2} – 4ac =  15



∴ x + 3 = 0 OR x – 2 = 0

∴ b^{2} – 4ac < 0



∴ x =  3 OR x =
2

∴ the roots are not real but imaginary numbers




∴ we can avoid this equation

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