1.
$${x}^{2}-5x+3=0$$
Solution
2.
$$-{x}^{2}+6x-8=3x+7$$
Solution
3.
$${2x}^{2}-x-1=0$$
Solution
4.
$$\frac{1}{2}{x}^{2}-16x=5$$
Solution
5.
$$\sqrt{3}{x}^{2}+\sqrt{5}x=12$$
Solution
6.
$$2{x}^{2}-67=-3$$
Solution
7.
$$3{x}^{2}+8x=0$$
Solution

A quadratic equation is any equation in the form
$$a{x}^{2}+bx+c=0$$
, where x is the unknown, and a, b, and c are known numbers, with a ≠ 0. The unknowns
$$a$$
,
$$b$$
and
$$c$$
are the coefficients of the equation and are called respectively, the quadratic coefficient, the linear coefficient and the constant term.
Note that although
$$a$$
must not be zero,
$$b$$
or
$$c$$
could be. This means, some quadratic equations might be missing the linear coefficient or the constant term, but they are perfectly valid. For example,
$$2{x}^{2}-64=0$$
lacks the linear coefficient (b=0), while
$$3{x}^{2}+8x=0$$
is missing the constant term (c=0).
There are many ways to solve quadratic equations, such as through factoring, completing the square, or using the quadratic formula. In the following section, we will demonstrate using the quadratic formula.

For a quadratic equation, which has the form
$$a{x}^{2}+bx+c=0$$
, its solutions are described by the quadratic formula. In other words, the values of
$$x$$
$$x=\frac{-b \pm \sqrt{{b}^{2}-4ac}}{2a}$$
In case you are not familiar with the
$$\pm$$
symbol, it indicates that, the expression on the right side of the equation can be expanded into two expressions, one with "
$$+$$
" and one with "
$$-$$
". Therefore, there are in fact two solutions to the quadratic equation. To be explicit, they are as follows:
$$x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}$$
and
$$x=\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$$

For practice, let’s solve the following quadratic equation using the quadratic formula.
$${x}^{2}+6x-8=3x+7$$
First, Let's move all the terms to one side, which gives:
$${x}^{2}+6x-8-3x-7=0$$
Then, simplify the equation again by combining terms:
$${x}^{2}+3x-15=0$$
Now, we have a quadratic equation that follows the form:
$$a{x}^{2}+bx+c=0$$
This means we can use the quadratic formula because we see that
$$a=1$$
,
$$b=3$$
and
$$c=-15$$
. Substituting these values into the quadratic equation yields:
$$x=\frac{-3 \pm \sqrt{{3}^{2}-(4)(-15)}}{2}$$
After simplification, we have:
$$x=\frac{-3 \pm \sqrt{69}}{2}$$
We are done! We have found the solutions (values of x) of the original equation.

# What's Next

Want to get better at solving quadratic equations and using the quadratic formula? Start with our practice problems at the top of this page and see if you can solve them. If you run into trouble, try the Cymath quadratic equation calculator to get the full solution and see the steps.
At Cymath, it is our goal to help students get better at math. Ready to take your learning to the next level with “how” and “why” steps? Sign up for Cymath Plus today.