Question

solve the quadratic equation numberically (using tables or x- and y- values ). (x+2)(x+3)=12 a.x=-1 or x=3 b. x=2 or x=-3 c. x=-6 or x=1 d. x=1 or x=-1

Answer 1

c.

Explanation:

`(x + 2)(x + 3) = 12 \\ {x}^(2) + 5x + 6 = 12 \\ {x}^(2) + 5x - 6 = 0 \\ (x + 6)(x - 1) = 0 \\ x = - 6 \: or \: x = 1`