Question

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

Answer 1

See a solution process below:

Explanation:

From: http://www.purplemath.com/modules/quadform.htm

The quadratic formula states:

For `ax^2 + bx + c = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

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

`x = (-(-31) +- sqrt((-31)^2 - (4 * 10 * 15)))/(2 * 10)`

`x = (31 +- sqrt(961 - 600))/20`

`x = (31 +- sqrt(361))/20`

`x = (31 +- 19)/20`

`x = 50/20` and `x = 12/20`

`x = 5/2` and `x = 3/5`

Answer 2

`5/2, 6/5`

Explanation:

Use the improved quadratic formula (Google, Socratic Search).
`f(x) = 10x^2 - 31x + 15 = 0`
`D = d^2 = b^2 - 4ac = 961 - 600 = 361` --> `d = +- 19`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 31/20 +- 19/20 = (31 +- 19)/20`
`x1 = 50/20 = 5/2`
`x2 = 12/20 = 3/5`