(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous
Step-by-step explanation:
Let the first integer be x
2nd integer = x + 1
3rd integer = x + 2
x + x + 1 + x + 2 = -147
3x + 3 = -147
3x = -147 - 3
3x = -150
x = -150 ÷ 3
x = -50
The three consecutive integers are
-50, -49, -48