Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

x^{2} + 5x – (a^{2} + a – 6) = 0

Given: x^{2} + 5x – (a^{2} + a – 6) = 0

Comparing with standard quadratic equation Ax^{2} + Bx + C = 0

A = 1, B = 5, C = – (a^{2} + a – 6)

Discriminant D = B^{2} – 4AC

= (5)^{2} – 4.1. – (a^{2} + a – 6)

= 25+ 4a^{2} + 4a – 24 = 4a^{2} + 4a + 1

= (2a + 1)^{2} > 0 Using a^{2} + 2ab + b^{2} = (a + b)^{2}

Hence the roots of equation are real.

= (2a + 1)

Roots are given by

x = (a – 2) or x = – (a + 3)

Hence the roots of equation are (a – 2) or x = – (a + 3)

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