A standard form for a quadratic equation is
$$ax^2+bx+c=0 \quad , a \neq 0$$
where $$a$$, $$b$$ and $$c$$ are constants.

$$x = \dfrac{- b\pm\sqrt{b^2-4a c}}{2a}$$

* Notice that $$x$$ has 2 values! So we get either
$$x = \dfrac{- b{\color{red}{+}}\sqrt{b^2-4a c}}{2a} \quad$$

or $$\quad x = \dfrac{- b{\color{red}{-}}\sqrt{b^2-4a c}}{2a}$$

## Special Cases

1 Case 1
Normally a quadratic formula has 2 values. However,  if $$b^2-4ac =0$$ , $$x$$ has only 1 value.

$$x = \dfrac{- b\pm\sqrt{b^2-4ac}}{2a} \\ \, \\ \quad \downarrow \\ \, \\ x = \dfrac{-b\pm\sqrt{0}}{2a} \\ \, \\ \quad \downarrow \\ \, \\ x = \dfrac{-b}{2a}$$

2 Case 2

If $$b^2-4ac$$ < 0, $$x$$ has no value.
$$\\ \, \\ x = \dfrac{- b\pm\sqrt{b^2-4ac}}{2a}$$

$$\quad \downarrow \\ \, \\ x = \dfrac{- b\pm\sqrt{\text{negative}}}{2a} \\ \, \\ \quad \downarrow$$
No value!

Example Solve $$2x^2+4x+1$$.

Solution  Using the quadratic formula, we can see that

$$a=2$$, $$b=4$$, $$c=1$$

$$\require{cancel} x = \dfrac{- b\pm\sqrt{b^2-4a c}}{2a} \\ \, \\ x = \dfrac{- 4\pm\sqrt{4^2-4(2)(1)}}{2(2)} \\ \, \\ x = \dfrac{- 4\pm\sqrt{16-8}}{4} \\ \, \\ x = \dfrac{- 4\pm\sqrt{8}}{4} \\ \, \\ x = \dfrac{- \cancel{{\color{red}{4}}}\pm\sqrt{8}}{\cancel{{\color{red}{4}}}} \\ \, \\ x = \pm \sqrt{8} \\ \, \\ x = \pm \sqrt{4 \times 2} \\ \, \\ x = \pm 2\sqrt{2} \\ \, \\$$
So we have either $$x= 2\sqrt{2}$$
or $$x= -2\sqrt{2}$$.

Solve $$x^2+5x-2=0$$ using the quadratic formula.
First, identify the values of $$a$$, $$b$$ and $$c$$.
$$a=1$$, $$b=5$$, $$c=-2$$
$$x = \dfrac{- b\pm\sqrt{b^2-4a c}}{2a} \\ \, \\ x = \dfrac{- 5\pm\sqrt{5^2-4(1)(-2)}}{2(1)} \\ \, \\ x = \dfrac{- 5\pm\sqrt{25+8}}{2} \\ \, \\ x = \dfrac{- 5\pm\sqrt{33}}{2} \\ \, \\$$
So we have either $$x = \dfrac{- 5+\sqrt{33}}{2}$$ or $$x = \dfrac{- 5-\sqrt{33}}{2}$$.