Solve (4+x)*(4-x)=33

$$(4+x) \times (4-x)=33$$

Solve for x (complex solution)

$$x=-\sqrt{17}i\approx -0-4.123105626i$$

$$x=\sqrt{17}i\approx 4.123105626i$$

Steps by Finding Square Root

Consider \left(4+x\right)\left(4-x\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 4.

$$16-x^{2}=33$$

Subtract 16 from both sides.

$$-x^{2}=33-16$$

Subtract 16 from 33 to get 17.

$$-x^{2}=17$$

Divide both sides by -1.

$$x^{2}=-17$$

The equation is now solved.

$$x=\sqrt{17}i x=-\sqrt{17}i$$

Steps Using the Quadratic Formula

Consider \left(4+x\right)\left(4-x\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 4.

$$16-x^{2}=33$$

Subtract 33 from both sides.

$$16-x^{2}-33=0$$

Subtract 33 from 16 to get -17.

$$-17-x^{2}=0$$

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.

$$-x^{2}-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\left(-17\right)}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\left(-17\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\left(-17\right)}}{2\left(-1\right)}$$

Multiply 4 times -17.

$$x=\frac{0±\sqrt{-68}}{2\left(-1\right)}$$

Take the square root of -68.

$$x=\frac{0±2\sqrt{17}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±2\sqrt{17}i}{-2}$$

Now solve the equation x=\frac{0±2\sqrt{17}i}{-2} when ± is plus.

$$x=-\sqrt{17}i$$

Now solve the equation x=\frac{0±2\sqrt{17}i}{-2} when ± is minus.

$$x=\sqrt{17}i$$

The equation is now solved.

$$x=-\sqrt{17}i x=\sqrt{17}i$$