So we are in the midst of solving quadratic equations of the form:

\[{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

where *a*,*b*, and *c* are some numbers. So far, we have been looking at equations like this that can be factored so that we can use the Null Factor Law. But most quadratics cannot be factored by hand and the solutions to the equations are frequently decimal numbers. So what to do? Well fortunately mathematicians have heard your concerns and way back in the year 628, an Indian mathematician Brahmagupta came up with a formula called the quadratic formula. This formula will solve any quadratic equation.

So given a quadratic equation, if you identify the three coefficients *a*,*b*, and *c*, the solution can be obtained by using the following formula:

{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}

\]

This is a very powerful equation. It is not difficult to prove but I’d rather you research that yourself and for now, just accept my word for it truthfulness. Let’s use it in some examples.

\[{2}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

Here *a* = 2, *b* = 5, and *c* = 3. Let’s put these numbers in the quadratic formula. Notice that there is a symbol ± in the formula. That means that there are generally two solutions: one using the + symbol and the other using the – symbol. So the solutions to this equation are:

{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{5}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{2}{)(}{3}{)}}}{2(2)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}{1}}{4}\hspace{0.33em}{=}\hspace{0.33em}{-}{1}{,}\hspace{0.33em}{-}{1}{.}{5}

\]

So there are two solutions to this quadratic equation. Let’s do another one:

\[{3}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{0}

\]

Here *a* = 3, *b* = -4, and *c* = -7. It’s important to include the signs in the *a*,*b*, and *c *and include them in the quadratic formula.

So the solutions are:

\[{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{(}{-}{4}{)}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{(}{-}{4}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{3}{)(}{-}{7}{)}}}{2(3)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{4}\hspace{0.33em}\pm\hspace{0.33em}{10}}{6}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{33}{,}\hspace{0.33em}{-}{1}

\]

Now look at the \[

{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}

\] part of the formula. If this is a positive number, like in the last two examples, then you can take the square root and get two solutions: one by using the + symbol and the other by using the – symbol. If this is zero, then you only get one solution as adding or subtracting a zero doesn’t give two solutions, just one. And if the expression is negative, well there is no solution in the real world because you can’t take the square root of a negative number. However, scientists and engineers work in a complex world where you can take the square root of a negative number, and this has a physical meaning. Maybe some day I will talk about these special numbers.