Let two consecutive number are x and (x+1)A.T.Qx(x+1)=306X2+x-306=0 X2+18x-17x-306=0x(x+18)-17(x+18)=0 (x-17) (x+18)=0 x-17=0 x+18=0x=17. x= -18 So ,x=17 and x+1 =17+1=18 Therefore,the no. are 17 and 18

First consecutive positive no is x and another x+1 so product x(x+1)=360
So x square + x+1 whole square =306And factors are +18and -17So x= 17 x+1 = 17+1=18
We have to let the consecutive positive integers be x and x+1So x(x+1)=306
Let two consecutive positive integers =x,x+1Product of two consecutive integers=306X(x+1)=306X square+x-306=0By quadratic formula-b+-√d\\2a-1+-√1-4(1)(-306)/2-1+-√1225/2-1+-35/2X=-1+35/2=34/2=17Or x=-36/2=-18So x=17So two consecutive positive integers=17and17+1=18

Let the 1st number be x . So the 2nd number be x+1. Acc. to question , (x)(x+1)=306 After solving this we get x^2 + x = 306. Then x^2 + x – 306 = 0 . Then by middle term factorization, we get x^2 + 17x – 18x – 306 = 0. Then x(x + 17) – 18(x + 17). Then (x – 18) (x – 17) = 0. Then x = 17 and x= 18. Therefore the integers are 17 and 18
17 and 18
X(x+1)=306X^2+x-306=0Simplyfy the qudratic equation and u will get the answer as 17 and 18

Let two consecutive numbers are x and (x + 1)A/C to question,product of x and (x + 1) = 306⇒x(x + 1) = 306⇒x² + x – 306 = 0⇒ x² + 18x – 17x – 306 = 0⇒x(x + 18) – 17(x + 18) = 0⇒(x + 18)(x – 17) = 0⇒ x = 17 and -18Hence, numbers are x = 17 and (x +1) = 18
17 and -18