Solve for x x^2-5x+9=0

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Trigonometry

$x^{2} - 5 x + 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 5$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Multiply $9$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 9}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$x = \frac{5 \pm \sqrt{25 - 36}}{2 \cdot 1}$

Subtract $36$ from $25$ .

$x = \frac{5 \pm \sqrt{- 11}}{2 \cdot 1}$

Rewrite $- 11$ as $- 1 (11)$ .

$x = \frac{5 \pm \sqrt{- 1 \cdot 11}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (11)}$ as $\sqrt{- 1} \cdot \sqrt{11}$ .

$x = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{11}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{5 \pm i \sqrt{11}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{5 \pm i \sqrt{11}}{2}$

The final answer is the combination of both solutions.

$x = \frac{5 + i \sqrt{11}}{2}, \frac{5 - i \sqrt{11}}{2}$

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