Find All Complex Solutions d^2+d-7=-5

$d^{2} + d - 7 = - 5$

Step 1

Move all terms to the left side of the equation and simplify.

Add $5$ to both sides of the equation.

$d^{2} + d - 7 + 5 = 0$

Add $- 7$ and $5$ .

$d^{2} + d - 2 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 1$ , and $c = - 2$ into the quadratic formula and solve for $d$ .

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

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$d = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

Multiply $- 4$ by $1$ .

$d = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$d = \frac{- 1 \pm \sqrt{1 + 8}}{2 \cdot 1}$

Add $1$ and $8$ .

$d = \frac{- 1 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$d = \frac{- 1 \pm \sqrt{3^{2}}}{2 \cdot 1}$

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

$d = \frac{- 1 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$d = \frac{- 1 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$d = 1, - 2$