Question

Three consecutive positive integers are such that the sum of the square of the first and the product od the two other is 46

Answer 1

let the first be x

then the second will be x+1

and the third will be x+2

according to the question

x2 + (x+1)(x+2) = 46

x2+x2+2x+2+x=46

2x2+3x+2=46

2x2+3x=44

3x=44-2x2

3x=2(22-x2)

3x/2=22-x2

Answer 2

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hello Friend}}}$


Here is your answer
Let the consecutive positive integer are x , x+1 , x+2.
ATQ
$x {}^{2} + (x + 1)(x + 2) = 46 \\ x {}^{2} + (x {}^{2} + 2x + x + 2) = 46 \\ x {}^{2} + x {}^{2} + 3x + 2 = 46 \\ 2x {}^{2} + 3x + 2 - 46 = 0 \\ 2x {}^{2} + 3x - 44 = 0 \\ 2x {}^{2} - 8x + 11x - 44 = 0 \\ 2x(x - 4) + 11(x - 4) = 0 \\ (x - 4)(2x + 11) = 0 \\ x - 4 = 0 \: \: \: and \: 2x + 11 = 0 \\ x = 4 \: \: and \: x = - \frac{11}{2} \\ as \: x \: is \: not \: a \: negaive \: number \\ so \: x = 4 \\ hence \: \: consecutive \:positive \: integer \: are \\ 4 \: \: . 5\: and \: 6$
Hope it helps you