X²+6x-(a²+2a-8)=0

The given equation is {tex}x^2 + 6x - (a^2 + 2a - 8) = 0{/tex}
Comparing it with {tex}Ax^2 + Bx +C = 0,{/tex} we get
{tex}A = 1,\  B = 6\ and\ C = -(a^2 + 2a - 8){/tex}
{tex}\therefore{/tex} {tex}D =B^2 - 4AC ={/tex} (6)² - 4(1)(-(a² + 2a - 8))

= 36 +4a² + 8a - 32

= 4a² + 8a + 4 = 4(a² + 2a + 1) = 4(a + 1)² > 0
So, the given equation has real roots, given by
{tex}\alpha = \frac { - B + \sqrt { D } } { 2 A } = \frac { - 6 + \sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \times 1 } = \frac { - 6 + 2 ( a + 1 ) } { 2 }{/tex} {tex}= -3 + (a + 1) = a - 2{/tex}
{tex}\beta = \frac { - B - \sqrt { D } } { 2 A } = \frac { - 6 - \sqrt { 4 ( a + 1 ) ^ { 2 } } } { 2 \times 1 } = \frac { - 6 - 2 ( a + 1 ) } { 2 }{/tex} {tex}= -3 - (a + 1) = -a - 4 = -(a + 4){/tex}
Hence, (a - 2) and -(a + 4) are the roots of the given equation.