Solve d=sqrt{12-d}

$$d = \sqrt { 12 - d }$$

Solve for d

$$d=3$$

Solution Steps

Square both sides of the equation.

$$d^{2}=\left(\sqrt{12-d}\right)^{2}$$

Calculate \sqrt{12-d} to the power of 2 and get 12-d.

$$d^{2}=12-d$$

Subtract 12 from both sides.

$$d^{2}-12=-d$$

Add d to both sides.

$$d^{2}-12+d=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$d^{2}+d-12=0$$

To solve the equation, factor d^{2}+d-12 using formula d^{2}+\left(a+b\right)d+ab=\left(d+a\right)\left(d+b\right). To find a and b, set up a system to be solved.

$$a+b=1 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

Rewrite factored expression \left(d+a\right)\left(d+b\right) using the obtained values.

$$\left(d-3\right)\left(d+4\right)$$

To find equation solutions, solve d-3=0 and d+4=0.

$$d=3 d=-4$$

Substitute 3 for d in the equation d=\sqrt{12-d}.

$$3=\sqrt{12-3}$$

Simplify. The value d=3 satisfies the equation.

$$3=3$$

Substitute -4 for d in the equation d=\sqrt{12-d}.

$$-4=\sqrt{12-\left(-4\right)}$$

Simplify. The value d=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-4=4$$

Equation d=\sqrt{12-d} has a unique solution.

$$d=3$$