How do you solve 3x^2+13x-10=0?

Jul 22, 2017

See a solution process below:

Explanation:

We can use the quadratic formula to solve this problem.

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:

$\textcolor{red}{3}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{13}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 10}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{{\textcolor{b l u e}{13}}^{2} - \left(4 \cdot \textcolor{red}{3} \cdot \textcolor{g r e e n}{- 10}\right)}}{2 \cdot \textcolor{red}{3}}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{169 - \left(- 120\right)}}{6}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{169 + 120}}{6}$

$x = \frac{- \textcolor{b l u e}{13} \pm \sqrt{289}}{6}$

$x = \frac{- \textcolor{b l u e}{13} + 17}{6}$ and $x = \frac{- \textcolor{b l u e}{13} - 17}{6}$

$x = \frac{4}{6}$ and $x = - \frac{30}{6}$

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

Jul 23, 2017

$\frac{2}{3}$ and - 5

Explanation:

$y = 3 {x}^{2} + 13 x - 10 = 0$
Use the new Transforming Method (Google Search):
Transformed equation:
$y ' = {x}^{2} + 13 x - 30 = 0$
Proceeding. Find 2 real roots of y', then. divide them by a = 3.
Find 2 real roots knowing the sum (-b = -13) and the product (ac = -30). They are: 2 and - 15.
Back to y, the 2 real roots are: $x 1 = \frac{2}{a} = \frac{2}{3}$ and
$x 2 = - \frac{15}{a} = - \frac{15}{3} = - 5$

Jul 23, 2017

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

Explanation:

Given:

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

Use an AC method to factor the quadratic:

Look for a pair of factors of $A C = 3 \cdot 10 = 30$ with difference $B = 13$

The pair $15 , 2$ works.

Use this pair to split the middle term and factor by grouping:

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

$\textcolor{w h i t e}{0} = \left(3 {x}^{2} + 15 x\right) - \left(2 x + 10\right)$

$\textcolor{w h i t e}{0} = 3 x \left(x + 5\right) - 2 \left(x + 5\right)$

$\textcolor{w h i t e}{0} = \left(3 x - 2\right) \left(x + 5\right)$

Hence zeros:

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