# What is the solution set for 2x^2 + 4x +10 = 0?

Aug 26, 2015

There are no Real solutions for the given equation.

#### Explanation:

We can see that there are no Real solutions by checking the discriminant
$\textcolor{w h i t e}{\text{XXX}} {b}^{2} - 4 a c$

#color(white)("XXX")= 16 - 80 < 0 color(white)("XX")rarrcolor(white)("XX")no Real roots

or
If we look at the graph for the expression, we can see that it does not cross the X-axis and therefore is not equal to zero at any values for $x$:
graph{2x^2+4x+10 [-10, 10, -5, 5]}

Aug 26, 2015

${x}_{1 , 2} = \frac{- 1 \pm 4 i}{2}$

#### Explanation:

For a general form quadratic equation

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

you can determine its roots by using the quadratic formula

$\textcolor{b l u e}{{x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$

Now, you can divide all the terms by $2$ to make the calculations easier

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} + \left(\frac{4}{2}\right) x + \frac{10}{2} = 0$

${x}^{2} + 2 x + 5 = 0$

For this quadratic, you have $a = 1$, $b = 2$, and $c = 5$, which means that the two roots will be

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{{2}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Notice that the determinant, $\Delta$, which is the name given to the expression that's under the square root, , is negative.

$\Delta = {b}^{2} - 4 a c$

$\Delta = {2}^{2} - 4 \cdot 1 \cdot 5 = - 16$

For real numbers, you cannot take the square root of a negative number, which means that the quadratic equation has no real solutions.

Its graph will not intercept the $x$-axis. However, it will have two distinct complex roots.

${x}_{1 , 2} = \frac{- 1 \pm \sqrt{- 16}}{2}$

${x}_{1 , 2} = \frac{- 1 \pm \left({i}^{2} \cdot 16\right)}{2} = \frac{- 1 \pm i \cdot \sqrt{16}}{2}$

${x}_{1 , 2} = \frac{- 1 \pm 4 i}{2}$

The two roots will thus be

${x}_{1} = \frac{- 1 + 4 i}{2} \text{ }$ and $\text{ } {x}_{2} = \frac{- 1 - 4 i}{2}$