Solve Using the Quadratic Formula (x+2)^2+(x-1)^2=0

${(x + 2)}^{2} + {(x - 1)}^{2} = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 2$ , $b = 2$ , and $c = 5$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot 5$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot 5}}{2 \cdot 2}$

Multiply $- 8$ by $5$ .

$x = \frac{- 2 \pm \sqrt{4 - 40}}{2 \cdot 2}$

Subtract $40$ from $4$ .

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

Rewrite $- 36$ as $- 1 (36)$ .

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

Rewrite $\sqrt{- 1 (36)}$ as $\sqrt{- 1} \cdot \sqrt{36}$ .

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

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

$x = \frac{- 2 \pm i \cdot \sqrt{36}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 2 \pm i \cdot \sqrt{6^{2}}}{2 \cdot 2}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm i \cdot 6}{2 \cdot 2}$

Move $6$ to the left of $i$ .

$x = \frac{- 2 \pm 6 i}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 6 i}{4}$

Simplify $\frac{- 2 \pm 6 i}{4}$ .

$x = \frac{- 1 \pm 3 i}{2}$