# How do you solve 11v ^ { 2} - 7v + 19= 9?

Nov 23, 2017

$v = \frac{7 \pm \sqrt{- 391}}{22}$

#### Explanation:

$11 {v}^{2} - 7 v + 19 = 9$

The 1st thing we do is to get $0$ on one side of the equation so that we can try to factor it (or use quadratic formula if not possible to factor). So we move the 9 to the other side of the equation.

$11 {v}^{2} - 7 v + 10 = 0$

This equation is in standard form, or $a {x}^{2} + b x + c = 0$

Now we can try to factor it. We have to find two numbers that multiply up to $a \cdot c$ (or $11 \cdot 10$) AND add up to $b$, or $- 7$.

So 2 numbers that multiply to $110$ and add up to $- 7$. There's NO number that does so!

So in this case we have to use the quadratic formula.
$v = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
We know that $a = 11$, $b = - 7$, and $c = 10$, so let's put these numbers in the formula.

$v = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(11\right) \left(10\right)}}{2 \left(11\right)}$
$v = \frac{7 \pm \sqrt{49 - 4 \left(110\right)}}{22}$
$v = \frac{7 \pm \sqrt{49 - 440}}{22}$
$v = \frac{7 \pm \sqrt{- 391}}{22}$

Nov 23, 2017

x = $\frac{7 + \sqrt{489}}{22}$ , x = $\frac{7 - \sqrt{489}}{22}$

#### Explanation:

Given:
$11 {v}^{2} - 7 v + 19 = 9$
Next, we subtract the coefficients;
$11 {v}^{2} - 7 v + 19 - 9 = 0$
$11 {v}^{2} - 7 v + 10 = 0$

We proceed to solve this question using the formula:

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

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

Substituting the values of a,b and c from $11 {v}^{2} - 7 v + 10 = 0$;
$\frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \cdot \left(11\right) \cdot \left(10\right)}}{2 \cdot 11}$
$\frac{\left(7\right) \pm \sqrt{- 391}}{22}$

Therefore, the values of x are:
x = $\frac{7 + \sqrt{391}}{22} i$
x = $\frac{7 - \sqrt{391}}{22} i$

Where i denotes a complex number.