Step-by-step explanation:

x

2

+4x−(a

2

+2a−3)=0

\implies \sf{ {x}^{2} + 4x - ( {a}^{2} + 3a - a- 3) \: } = 0⟹x

2

+4x−(a

2

+3a−a−3)=0

\implies \sf{ {x}^{2} + 4x - \bigg [ {a}(a + 3) - 1(a + 3) \bigg] \: } = 0⟹x

2

+4x−[a(a+3)−1(a+3)]=0

\implies \sf{ {x}^{2} + 4x - (a + 3) (a - 1) \: } = 0⟹x

2

+4x−(a+3)(a−1)=0

\implies \sf{ {x}^{2} + \bigg[ (a + 3) - (a - 1) \bigg] x - (a + 3) (a - 1) \: } = 0⟹x

2

+[(a+3)−(a−1)]x−(a+3)(a−1)=0

\implies \sf{ {x}^{2} + (a + 3) x- (a - 1) x - (a + 3) (a - 1) \: } = 0⟹x

2

+(a+3)x−(a−1)x−(a+3)(a−1)=0

\implies \sf{ {x}(x+ a + 3) - (a - 1) (x + a + 3) \: } = 0⟹x(x+a+3)−(a−1)(x+a+3)=0

\implies \sf{ (x+ a + 3) (x - a + 1)\: } = 0⟹(x+a+3)(x−a+1)=0

Since the product of two real numbers are zero then either of them are zero

\implies \sf{ (x+ a + 3) = 0 \: \: \: or \: \: \: (x - a + 1)\: } = 0⟹(x+a+3)=0or(x−a+1)=0

Now

\sf{ (x+ a + 3) = 0 \:} \: \: \: gives(x+a+3)=0gives

\sf{ x = - ( a + 3) \:} \: \: \:x=−(a+3)

Again

\sf{ (x - a + 1) = 0 \:} \: \: \: gives(x−a+1)=0gives

\sf{ x = a - 1 \:} \: \: \:x=a−1

Hence the required solution is

\sf{x = - (a+3 ) \: \: , \: \: (a-1)}x=−(a+3),(a−1)