# How do you find the discriminant and how many and what type of solutions does x^2 + 10x + 25 = 64 have?

May 8, 2018

The quadratic has two real solutions. (The solutions are $x = - 13$ and $x = 3$.)

#### Explanation:

To find the discriminant, first bring everything to one side of the equation:

${x}^{2} + 10 x + 25 = 64$

${x}^{2} + 10 x - 39 = 0$

Now, use the $a$, $b$, and $c$ values ($1$, $10$, and $- 39$) and plug them into the discriminant of the quadratic formula (highlighted in red):

$x = \frac{- b \pm \sqrt{\textcolor{red}{{b}^{2} - 4 a c}}}{2 a}$

Discriminant:

$\textcolor{w h i t e}{=} {b}^{2} - 4 a c$

$= {\left(10\right)}^{2} - 4 \left(1\right) \left(- 39\right)$

$= 100 + 156$

$= 256$

Since the discriminant is positive, the quadratic has two real solutions.

To solve for the solutions, you can use the quadratic formula:

$x = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \left(1\right) \left(39\right)}}{2 \left(1\right)}$

$x = \frac{- 10 \pm 16}{2}$

$x = - 5 \pm 8$

$x = - 13 , 3$

Or you could factor the quadratic:

${x}^{2} + 10 x - 39 = 0$

$\left(x + 13\right) \left(x - 3\right) = 0$

$x = - 13 , 3$

Or you could take the original problem and square root both sides:

${x}^{2} + 10 x + 25 = 64$

$\left(x + 5\right) \left(x + 5\right) = 64$

${\left(x + 5\right)}^{2} = 64$

$x + 5 = \pm \sqrt{64}$

$x + 5 = \pm 8$

$x = \pm 8 - 5$

$x = - 13 , 3$

Hope this helped!