[tex] {x}^{2} + 4x - 12 = 0 \\ {x}^{2} + 4x + 4 - 4 - 12 = 0\\ {(x + 2)}^{2} - 16 = 0 \\ x = - + \sqrt{16} - 2 \\ x = + \: or - 4 - 2 \\ x = - 6 \: \: x = 2[/tex]
- subtract the 12 from both sides so that it becomes the last constant term in the quadratic equation which should now equal 0.
- take the 4x
- half the coefficient of 4 (2)
- square it (4)
- add it to the equation (+4)
- subtract it from the equation (-4)
- factorise the square (x+2)^2 expands to (x^2 + 4x + 2) as {a+b}^2={a^2 + ab + ba + b^2}
- now the equation is in turning point form.
- to find x, add 16 and square root 16 and (x+2)
- subtract 2 from positive or negative 4 (as -4^2 and 4^2 both equal 16).
- This should give you two values for x, -6 and 2.
I really hope that this helped :)
