x²+x-(a+1)(a+2) = 0
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Hey there!
We have ,
x^2 + x - (a+1)(a+2) = 0
x^2 + x {(a+2 ) -(a+1) } - (a+1)(a+2) =0
{ x^2 + x (a+2)} -x (a +1) -(a + 1) (a+2)=0
x { x +(a+2) } - (a+1) { x+ (a+2)} =0
{x + (a+2) } { x - (a+1)} =0
x + (a +2) = 0 or , x - (a+1) =0
So, x = -(a+2) or, x = (a+1)
Hope it helps!
We have ,
x^2 + x - (a+1)(a+2) = 0
x^2 + x {(a+2 ) -(a+1) } - (a+1)(a+2) =0
{ x^2 + x (a+2)} -x (a +1) -(a + 1) (a+2)=0
x { x +(a+2) } - (a+1) { x+ (a+2)} =0
{x + (a+2) } { x - (a+1)} =0
x + (a +2) = 0 or , x - (a+1) =0
So, x = -(a+2) or, x = (a+1)
Hope it helps!
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