Find All Complex Solutions 2x^2+7x+k=0

$2 x^{2} + 7 x + k = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = 7$ , and $c = k$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (2 \cdot k)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot k}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot k}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{- 7 + \sqrt{49 - 8 k}}{4}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \cdot 7 + \sqrt{49 - 8 k}}{4}$

Factor $- 1$ out of $\sqrt{49 - 8 k}$ .

$x = \frac{- 1 \cdot 7 - 1 (- \sqrt{49 - 8 k})}{4}$

Factor $- 1$ out of $- 1 (7) - 1 (- \sqrt{49 - 8 k})$ .

$x = \frac{- 1 (7 - \sqrt{49 - 8 k})}{4}$

Move the negative in front of the fraction.

$x = - \frac{7 - \sqrt{49 - 8 k}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot k}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 k}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{- 7 - \sqrt{49 - 8 k}}{4}$

Factor $- 1$ out of $- 7 - \sqrt{49 - 8 k}$ .

Reorder $49$ and $- 8 k$ .

$x = \frac{- 7 - \sqrt{- 8 k + 49}}{4}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 1 \cdot 7 - \sqrt{- 8 k + 49}}{4}$

Factor $- 1$ out of $- \sqrt{- 8 k + 49}$ .

$x = \frac{- 1 \cdot 7 - (\sqrt{- 8 k + 49})}{4}$

Factor $- 1$ out of $- 1 (7) - (\sqrt{- 8 k + 49})$ .

$x = \frac{- 1 (7 + \sqrt{- 8 k + 49})}{4}$

Rewrite $- 1 (7 + \sqrt{- 8 k + 49})$ as $- (7 + \sqrt{- 8 k + 49})$ .

$x = \frac{- (7 + \sqrt{- 8 k + 49})}{4}$

Move the negative in front of the fraction.

$x = - \frac{7 + \sqrt{- 8 k + 49}}{4}$

The final answer is the combination of both solutions.

$x = - \frac{7 - \sqrt{49 - 8 k}}{4}$

$x = - \frac{7 + \sqrt{- 8 k + 49}}{4}$

Find All Complex Solutions 2x^2+7x+k=0

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