How do you solve $3x^2 + 13x = 10$ ?

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Answer 1 · 2018-03-26T01:29:09

See a solution process below:

First, subtract $color(red)(10)$ from each side of the equation to put the equation in standard quadratic for while keeping the equation balanced:

$3x^2 + 13x - color(red)(10) = 10 - color(red)(10)$

$3x^2 + 13x - 10 = 0$

We can now use the quadratic equation to solve this problem:

For $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$color(red)(3)$ for $color(red)(a)$

$color(blue)(13)$ for $color(blue)(b)$

$color(green)(-10)$ for $color(green)(c)$ gives:

$x = (-color(blue)(13) +- sqrt(color(blue)(13)^2 - (4 * color(red)(3) * color(green)(-10))))/(2 * color(red)(3))$

$x = (-color(blue)(13) +- sqrt(169 - (12 * color(green)(-10))))/6$

$x = (-color(blue)(13) +- sqrt(169 - (-120)))/6$

$x = (-color(blue)(13) +- sqrt(169 + 120))/6$

$x = (-color(blue)(13) +- sqrt(289))/6$

$x = (-color(blue)(13) - 17)/6$ and $x = (-color(blue)(13) + 17)/6$

$x = -30/6$ and $x = 4/6$

$x = -5$ and $x = 2/3$

The Solution Set Is: $x = {-5, 2/3}$

How do you solve 3x^2 + 13x = 103x^2 + 13x = 10?