# How do you solve x(x+6)=-2 using the quadratic formula?

Oct 29, 2017

$x = - 0.354$
OR
$x = - 5.645$

#### Explanation:

$x \left(x + 6\right) = - 2$

Writing it in standard form:

${x}^{2} + 6 x + 2 = 0$

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

$a = 1$
$b = 6$
$c = 2$

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

$x = - 0.354248688 \approx - 0.354$---truncated and approximate value.

Or

$x = - 5.645751311 \approx - 5.645$---truncated and approximate value.

$\therefore x = - 0.354$
OR
$x = - 5.645$

Oct 29, 2017

$x = - 3 + \sqrt[2]{7} \mathmr{and} \approx - 0.3542 \ldots$
$x = - 3 - \sqrt[2]{7} \mathmr{and} \approx - 5.6457 \ldots$

#### Explanation:

Expand and rearnage;
${x}^{2} + 6 x = - 2$
${x}^{2} + 6 x + 2 = 0$

Then apply to the quadratic formula;
$a = 1 , b = 6 , c = 2$

$x = \frac{- 6 + \sqrt[2]{{6}^{2} - \left(4 \cdot 1 \cdot 2\right)}}{2}$
or $x = \frac{- 6 - \sqrt[2]{{6}^{2} - \left(4 \cdot 1 \cdot 2\right)}}{2}$

To give
$x = \frac{- 6 + \sqrt[2]{28}}{2}$
or $x = \frac{- 6 - \sqrt[2]{28}}{2}$

Hence
$x = \frac{- 6 + 2 \sqrt[2]{7}}{2}$
or $x = \frac{- 6 - 2 \sqrt[2]{7}}{2}$

Finally giving;
$x = - 3 + \sqrt[2]{7} \mathmr{and} \approx - 0.3542 \ldots$
$x = - 3 - \sqrt[2]{7} \mathmr{and} \approx - 5.6457 \ldots$