# Solve by using the quadratic formula?

## $3 {x}^{2} + 4 x + 10 = 0$ Jun 18, 2018

See a solution process below:

#### Explanation:

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:

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

$\textcolor{b l u e}{4}$ for $\textcolor{b l u e}{b}$

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

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

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{16 - 120}}{6}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{- 104}}{6}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{4 \times - 26}}{6}$

$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{4} \sqrt{- 26}}{6}$

$x = \frac{- \textcolor{b l u e}{4} \pm 2 \sqrt{- 26}}{6}$

Jun 18, 2018

No real solution.

#### Explanation:

The quadratic formular is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ for the equation $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{\mathmr{and} a n \ge}{c} = 0$

Therefore, in your case ($\textcolor{red}{3} {x}^{2} + \textcolor{b l u e}{4} x + \textcolor{\mathmr{and} a n \ge}{10} = 0$)
$a = \textcolor{red}{3}$
$b = \textcolor{b l u e}{4}$
$c = \textcolor{\mathmr{and} a n \ge}{10}$

Using the formular, we get:
$x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{{\textcolor{b l u e}{4}}^{2} - 4 \cdot \textcolor{red}{3} \cdot \textcolor{\mathmr{and} a n \ge}{10}}}{2 \cdot \textcolor{red}{3}}$
$x = \frac{- 4 \pm \sqrt{16 - 120}}{6}$
$x = - \frac{2}{3} \pm \frac{\sqrt{\textcolor{g r e e n}{- 104}}}{6}$

Since the radicand ($\textcolor{g r e e n}{- 104}$) is negative, this equation has no real solutions for $x$.