How do you work out this Quadratic:: (x+3)(x-4)=18?

x² - x - 12 = 18
x² - x - 30 = 0
( x - 6 ) ( x + 5 ) = 0
x = 6 , x = - 5

(x + 3)(x - 4) = 18
x^2 + 3x - 4x - 12 = 18
x^2 - x - 12 - 18 = 0
x^2 - x - 30 = 0
(x - 6)(x + 5) = 0

x - 6 = 0
x = 6

x + 5 = 0
x = -5

∴ x = 6 , -5

(x+3)(x-4)=18
x^2+3x-4x-12=18
x^2-x-30=0
(x-6)(x+5)=0
x=6 or x=-5

hope that helped

you did ask how.

multiply out the left side. Move the 18 to the left and set equation =0. Combine terms. Use quadratic equation to solve.

First, multiply back to polynomial form:

x^2 -x -12 = 18

Then, subtract both sides by 18, so that you leave zero on the right side of the equal sign:

x^2 - x -30 = 0

Now, factor once again (reverse foil):

(x + 5)(x - 6) = 0

To satisfy :

x = -5 or x = 6

Plug back in to original equation to see if this solution set works!

You first multiply 4 and 3 which equals 12. Then you subtract it from 18 which makes 6. Then you so you are six short. Then you put these numbers in:
(3+3)(7-4)=18 and that is the answer.