How do you solve 10x^2-31x+15=0 using the quadratic formula?

Jun 12, 2017

See a solution process below:

Explanation:

The quadratic formula states:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting $10$ for $a$; $- 31$ for $b$ and $15$ for $c$ gives:

$x = \frac{- \left(- 31\right) \pm \sqrt{{\left(- 31\right)}^{2} - \left(4 \cdot 10 \cdot 15\right)}}{2 \cdot 10}$

$x = \frac{31 \pm \sqrt{961 - 600}}{20}$

$x = \frac{31 \pm \sqrt{361}}{20}$

$x = \frac{31 \pm 19}{20}$

$x = \frac{50}{20}$ and $x = \frac{12}{20}$

$x = \frac{5}{2}$ and $x = \frac{3}{5}$

Jun 12, 2017

$\frac{5}{2} , \frac{6}{5}$

Explanation:

Use the improved quadratic formula (Google, Socratic Search).
$f \left(x\right) = 10 {x}^{2} - 31 x + 15 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 961 - 600 = 361$ --> $d = \pm 19$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{31}{20} \pm \frac{19}{20} = \frac{31 \pm 19}{20}$
$x 1 = \frac{50}{20} = \frac{5}{2}$
$x 2 = \frac{12}{20} = \frac{3}{5}$