Can somebody help break this problem down for me using the quadratic formula? Thanks!
x^2+2x-2=0
I know the answer is x=-1 plusminus square root 3 (don't know how to write that on here)
For $\displaystyle ax^2+bx+c = 0$ then $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
If you compare your equation to $\displaystyle ax^2+bx+c=0$ then you'll see that $\displaystyle a=1$, $\displaystyle b=2$ and $\displaystyle c=-2$. The place most people split up is a sign error when calculating $\displaystyle b^2-4ac$
$\displaystyle x = \frac{-2 \pm \sqrt{4+8}}{2}$
$\displaystyle x = \frac{-2 \pm \sqrt{12}}{2}$
$\displaystyle x = \frac{-2 \pm 2\sqrt{3}}{2}$
so,
$\displaystyle x = \frac{-2 + 2\sqrt{3}}{2} = {\sqrt{3} -1}$
OR
$\displaystyle x = \frac{-2 - 2\sqrt{3}}{2}= {-\sqrt{3} -1}$
Both values of x satisfy your equation
$\displaystyle -2+2\sqrt{3} = 2\times -1+2\times \sqrt{3}$
Now 2 is common to both terms so it can be factored out giving
$\displaystyle 2\times -1+2\times \sqrt{3} = 2(-1+\sqrt{3})$
If you cannot follow this please revise basic factorisation before asking additional questions.