Solve 3(x+3)(x-3)-5x=(x-4)(5x+4)=

$$3 ( x + 3 ) ( x - 3 ) - 5 x = ( x - 4 ) ( 5 x + 4 ) =$$

Solve for x

$$x = \frac{\sqrt{33} + 11}{4} \approx 4.186140662$$

$$x = \frac{11 - \sqrt{33}}{4} \approx 1.313859338$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x+3.

$$\left(3x+9\right)\left(x-3\right)-5x=\left(x-4\right)\left(5x+4\right)$$

Use the distributive property to multiply 3x+9 by x-3 and combine like terms.

$$3x^{2}-27-5x=\left(x-4\right)\left(5x+4\right)$$

Use the distributive property to multiply x-4 by 5x+4 and combine like terms.

$$3x^{2}-27-5x=5x^{2}-16x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-27-5x-5x^{2}=-16x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-27-5x=-16x-16$$

Add 16x to both sides.

$$-2x^{2}-27-5x+16x=-16$$

Combine -5x and 16x to get 11x.

$$-2x^{2}-27+11x=-16$$

Add 16 to both sides.

$$-2x^{2}-27+11x+16=0$$

Add -27 and 16 to get -11.

$$-2x^{2}-11+11x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+11x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-11\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-11\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-11\right)}}{2\left(-2\right)}$$

Multiply 8 times -11.

$$x=\frac{-11±\sqrt{121-88}}{2\left(-2\right)}$$

Add 121 to -88.

$$x=\frac{-11±\sqrt{33}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±\sqrt{33}}{-4}$$

Now solve the equation x=\frac{-11±\sqrt{33}}{-4} when ± is plus. Add -11 to \sqrt{33}.

$$x=\frac{\sqrt{33}-11}{-4}$$

Divide -11+\sqrt{33} by -4.

$$x=\frac{11-\sqrt{33}}{4}$$

Now solve the equation x=\frac{-11±\sqrt{33}}{-4} when ± is minus. Subtract \sqrt{33} from -11.

$$x=\frac{-\sqrt{33}-11}{-4}$$

Divide -11-\sqrt{33} by -4.

$$x=\frac{\sqrt{33}+11}{4}$$

The equation is now solved.

$$x=\frac{11-\sqrt{33}}{4} x=\frac{\sqrt{33}+11}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 3 by x+3.

$$\left(3x+9\right)\left(x-3\right)-5x=\left(x-4\right)\left(5x+4\right)$$

Use the distributive property to multiply 3x+9 by x-3 and combine like terms.

$$3x^{2}-27-5x=\left(x-4\right)\left(5x+4\right)$$

Use the distributive property to multiply x-4 by 5x+4 and combine like terms.

$$3x^{2}-27-5x=5x^{2}-16x-16$$

Subtract 5x^{2} from both sides.

$$3x^{2}-27-5x-5x^{2}=-16x-16$$

Combine 3x^{2} and -5x^{2} to get -2x^{2}.

$$-2x^{2}-27-5x=-16x-16$$

Add 16x to both sides.

$$-2x^{2}-27-5x+16x=-16$$

Combine -5x and 16x to get 11x.

$$-2x^{2}-27+11x=-16$$

Add 27 to both sides.

$$-2x^{2}+11x=-16+27$$

Add -16 and 27 to get 11.

$$-2x^{2}+11x=11$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{11}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{11}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{11}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=-\frac{11}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{11}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{11}{2}+\frac{121}{16}$$

Add -\frac{11}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{33}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{\sqrt{33}}{4} x-\frac{11}{4}=-\frac{\sqrt{33}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{\sqrt{33}+11}{4} x=\frac{11-\sqrt{33}}{4}$$