how do you solve this quadratics word problem?

Question
how do you solve this quadratics word problem?
7.Find 3 consecutive integers such that the square of the first plus the product of the other two is 46.

i'm pretty sure the equation is x2 + yz = 46 but i don't know how to solve it.

ANSWER:
x = 1st integer
(x + 1) = 2nd integer
(x + 2) = 3rd integer

the square of the first plus the product of the other two is 46.
x^2 + (x + 1) (x + 2) = 46
x^2 + x^2 + x + 2x + 2 = 46
2x^2 + 3x + 2 - 46 = 0
2x^2 + 3x - 44 = 0
(2x - 11) (x + 4) = 0
2x - 11 = 0 (or) x + 4 = 0
2x = 11 (or) x = -4
x =11/2 (or) x = -4
x = 5.5 (or) x = -4
integer is -4

x + 1 = - 4 + 1 = -3
x + 2 = -4 + 2 = -2

-4 = 1st integer
-3 = 2nd integer
-2 = 3rd integer

I believe the equation should be x2 + 2x = 46 and than you move 46 to the other side and you get x2+2x-46=0 . From there you do the quadratic formula which is x = -b +- square root of b2 - 4 ac all over 2. well I cant really do it on the computer but it should b pretty easy. if this is wrong srry not that good with math lol.

x, x+1, x+2 are the integers

x² + (x+1)(x+2) = 46
x² + (x²+3x+2) = 46
2x² + 3x ? 44 = 0
(2x + 11)(x ? 4) = 0
x = ?11/2, 4

so the integers are 4,5,6
check
16 + 30 = 46
ok

.