Suppose we want to solve $ax^2 + bx + c = 0$, where
$a \neq 0$.

The general solution is given by the formula

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$

if $b^2 - 4ac \geq 0$. On the other hand, if $b^2-4ac\textless 0$, the quadratic equation has no real solution.

\begin{xmpl}

Solve $x^2 - 3x + 2 = 0$

Putting this into the context of the formulation $ax^2+bx+c=0$, the constants are;

$a = 1, b = -3 , c = 2$

Inserting this into the formula for the roots gives:

\begin{eqnarray*}
x &=& \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}\\
x &=& \frac{3 \pm \sqrt{9 - 8}}{2}\\
x &=& \frac{3 \pm \sqrt{1}}{2}\\
x &=& \frac{3 + 1}{2} , \frac{3 - 1}{2}\\
x &=& \frac{4}{2} , \frac{2}{2}\\
x &=& 2 , 1
\end{eqnarray*}

\end{xmpl}
\begin{xmpl}


Find the roots of the following polynomial

$$3x^{4} + 14x^{2} + 15$$

We can use the quadratic equation to solve for the roots of this polynomial if we substitute a variable for $$x^{2}$$ Let's use the letter $$a$$

$$3a^{2} + 14a + 15$$

We then plug the constants in to the quadratic equation.

$$x = \frac{-(14) \pm \sqrt{14^{2} - (4)(3)(15)}}{(2)(3)}$$

which simplifies to

$$\frac{-(14) \pm \sqrt{196 - 180}}{6}$$

which equals $$-1\frac{2}{3}$$ and $$-3$$.

Then, since we substituted a for $$x^2$$ we need to take the square root of these values to get the roots of the polynomial.

So, $$x_{1,2} = \pm \sqrt{-1\frac{2}{3}}$$ and $$x_{3,4} = \pm \sqrt{3}$$
\end{xmpl}