Solve for d d^2+4=6d

$d^{2} + 4 = 6 d$

Subtract $6 d$ from both sides of the equation.

$d^{2} + 4 - 6 d = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 4$ into the quadratic formula and solve for $d$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$d = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$d = \frac{6 \pm \sqrt{36 - 4 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $4$ .

$d = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 1}$

Subtract $16$ from $36$ .

$d = \frac{6 \pm \sqrt{20}}{2 \cdot 1}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$d = \frac{6 \pm \sqrt{4 (5)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$d = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$d = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$d = \frac{6 \pm 2 \sqrt{5}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{5}}{2}$ .

$d = 3 \pm \sqrt{5}$

The final answer is the combination of both solutions.

$d = 3 + \sqrt{5}, 3 - \sqrt{5}$

The result can be shown in multiple forms.

Exact Form:

$d = 3 + \sqrt{5}, 3 - \sqrt{5}$

Decimal Form:

$d = 5.23606797 \dots, 0.76393202 \dots$