Solve (3-d)(3+d)=8

$$( 3 - d ) ( 3 + d ) = 8$$

Solve for d

$$d=1$$

$$d=-1$$

Steps by Finding Square Root

Consider \left(3-d\right)\left(3+d\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9-d^{2}=8$$

Subtract 9 from both sides.

$$-d^{2}=8-9$$

Subtract 9 from 8 to get -1.

$$-d^{2}=-1$$

Divide both sides by -1.

$$d^{2}=\frac{-1}{-1}$$

Divide -1 by -1 to get 1.

$$d^{2}=1$$

Take the square root of both sides of the equation.

$$d=1 d=-1$$

Steps Using the Quadratic Formula

Consider \left(3-d\right)\left(3+d\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$9-d^{2}=8$$

Subtract 8 from both sides.

$$9-d^{2}-8=0$$

Subtract 8 from 9 to get 1.

$$1-d^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-d^{2}+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$d=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square 0.

$$d=\frac{0±\sqrt{-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$d=\frac{0±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$d=\frac{0±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$d=\frac{0±2}{-2}$$

Now solve the equation d=\frac{0±2}{-2} when ± is plus. Divide 2 by -2.

$$d=-1$$

Now solve the equation d=\frac{0±2}{-2} when ± is minus. Divide -2 by -2.

$$d=1$$

The equation is now solved.

$$d=-1 d=1$$