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

Oct 22, 2017

See a solution process below:

Explanation:

One solution is to factor the quadratic as:

$\left(x - 2\right) \left(x + 5\right) = 0$

Now, we can solve each term on the left for $0$ to find the solutions:

Solution 1:

$x - 2 = 0$

$x - 2 + \textcolor{red}{2} = 0 + \textcolor{red}{2}$

$x - 0 = 2$

$x = 2$

Solution 2:

$x + 5 = 0$

$x + 5 - \textcolor{red}{5} = 0 - \textcolor{red}{5}$

$x + 0 = - 5$

$x = - 5$

he Solutions Are: $x = 2$ and $x = - 5$

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

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{3}$ for $\textcolor{b l u e}{b}$

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

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{{\textcolor{b l u e}{3}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 10}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 - \left(- 40\right)}}{2}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 + 40}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - \sqrt{9 + 40}}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + \sqrt{9 + 40}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - \sqrt{49}}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + \sqrt{49}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - 7}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + 7}{2}$

$x = - \frac{10}{2}$ and $x = \frac{4}{2}$

$x = - 5$ and $x = 2$