Question

What are the zeros of the equation x^2+4x-12=0

Answer 1

The first method:
$x^2+4x-12=0\ \ \ |+12\\\\x^2+2\cdot x\cdot2=12\ \ \ \ |+2^2\\\\\underbrace{x^2+2\cdot x\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}=12+2^2\\\\(x+2)^2=16\to x+2=\pm\sqrt{16}\\\\x+2=-4\ \vee\ x+2=4\ \ \ |-2\\\\x=-6\ \vee\ x=2$

The second method:
$x^2+4x-12=0\\\\a=1;\ b=4;\ c=-12\\\\b^2-4ac=4^2-4\cdot1\cdot(-12)=16+48=64\\\\\sqrt{b^2-4ac}=\sqrt{64}=8\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-4-8}{2\cdot1}=\dfrac{-12}{2}=-6\\\\x_2=\dfrac{-4+8}{2\cdot1}=\dfrac{4}{2}=2$

Third method:
$x^2+4x-12=0\\\\x^2+6x-2x-12=0\\\\x(x+6)-2(x+6)=0\\\\(x+6)(x-2)=0\iff x+6=0\ \vee\ x-2=0\\\\x=-6\ \vee\ x=2$