# How do you solve 4x^2-x+5=0?

Nov 18, 2016

The answer is $= \frac{1 \pm i \sqrt{79}}{8}$

#### Explanation:

The equation is $a {x}^{2} + b x + c = 0$

$4 {x}^{2} - x + 5 = 0$

Let's calculate the discriminant,

$\Delta = {b}^{2} - 4 a c = 1 - 4 \cdot 4 \cdot 5 = - 79$

$\Delta < 0$

So the solutions are not in $\mathbb{R}$ but in $\mathbb{C}$

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$x = \frac{1 \pm i \sqrt{79}}{8}$

where ${i}^{2} = - 1$

graph{4x^2-x+5 [-24.46, 21.15, -2.57, 20.25]}

Nov 18, 2016

$x = \frac{1}{8} \setminus \pm \frac{i \sqrt{79}}{8}$

#### Explanation:

If you have an equation in the form of $a {x}^{2} + b x + c = 0$ you can apply the quadratic formula

$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where $a$ is the coefficient of ${x}^{2}$ and $b$ is the coefficient of $x$ and $c$ is the constant

$4 {x}^{2} - x + 5 = 0$

In this example $a = 4$ and $b = - 1$ and $c = 5$

$x = \frac{- \left(- 1\right) \setminus \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(4\right) \left(5\right)}}{2 \left(4\right)}$

$= \frac{1 \setminus \pm \sqrt{- 79}}{8} = \frac{1}{8} \setminus \pm \frac{\sqrt{- 79}}{8}$

You can not have a negative under the square root, we can use the definition that $i = \sqrt{- 1}$ to extract the negative

$\frac{1}{8} \setminus \pm \frac{\sqrt{- 79}}{8} = \frac{1}{8} \setminus \pm \frac{i \sqrt{79}}{8}$

The final answer is $x = \frac{1}{8} \setminus \pm \frac{i \sqrt{79}}{8}$