Solve 3x-4*(x-3)=9x*(x+1)

$$3x-4 \times (x-3)=9x \times (x+1)$$

Solve for x

$$x=\frac{\sqrt{133}-5}{9}\approx 0.725840288$$

$$x=\frac{-\sqrt{133}-5}{9}\approx -1.836951399$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4 by x-3.

$$3x-4x+12=9x\left(x+1\right)$$

Combine 3x and -4x to get -x.

$$-x+12=9x\left(x+1\right)$$

Use the distributive property to multiply 9x by x+1.

$$-x+12=9x^{2}+9x$$

Subtract 9x^{2} from both sides.

$$-x+12-9x^{2}=9x$$

Subtract 9x from both sides.

$$-x+12-9x^{2}-9x=0$$

Combine -x and -9x to get -10x.

$$-10x+12-9x^{2}=0$$

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.

$$-9x^{2}-10x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -10 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-9\right)\times 12}}{2\left(-9\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-9\right)\times 12}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-\left(-10\right)±\sqrt{100+36\times 12}}{2\left(-9\right)}$$

Multiply 36 times 12.

$$x=\frac{-\left(-10\right)±\sqrt{100+432}}{2\left(-9\right)}$$

Add 100 to 432.

$$x=\frac{-\left(-10\right)±\sqrt{532}}{2\left(-9\right)}$$

Take the square root of 532.

$$x=\frac{-\left(-10\right)±2\sqrt{133}}{2\left(-9\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{133}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{10±2\sqrt{133}}{-18}$$

Now solve the equation x=\frac{10±2\sqrt{133}}{-18} when ± is plus. Add 10 to 2\sqrt{133}.

$$x=\frac{2\sqrt{133}+10}{-18}$$

Divide 10+2\sqrt{133} by -18.

$$x=\frac{-\sqrt{133}-5}{9}$$

Now solve the equation x=\frac{10±2\sqrt{133}}{-18} when ± is minus. Subtract 2\sqrt{133} from 10.

$$x=\frac{10-2\sqrt{133}}{-18}$$

Divide 10-2\sqrt{133} by -18.

$$x=\frac{\sqrt{133}-5}{9}$$

The equation is now solved.

$$x=\frac{-\sqrt{133}-5}{9} x=\frac{\sqrt{133}-5}{9}$$

Steps for Completing the Square

Use the distributive property to multiply -4 by x-3.

$$3x-4x+12=9x\left(x+1\right)$$

Combine 3x and -4x to get -x.

$$-x+12=9x\left(x+1\right)$$

Use the distributive property to multiply 9x by x+1.

$$-x+12=9x^{2}+9x$$

Subtract 9x^{2} from both sides.

$$-x+12-9x^{2}=9x$$

Subtract 9x from both sides.

$$-x+12-9x^{2}-9x=0$$

Combine -x and -9x to get -10x.

$$-10x+12-9x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-10x-9x^{2}=-12$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-9x^{2}-10x=-12$$

Divide both sides by -9.

$$\frac{-9x^{2}-10x}{-9}=-\frac{12}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{10}{-9}\right)x=-\frac{12}{-9}$$

Divide -10 by -9.

$$x^{2}+\frac{10}{9}x=-\frac{12}{-9}$$

Reduce the fraction \frac{-12}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{9}x=\frac{4}{3}$$

Divide \frac{10}{9}, the coefficient of the x term, by 2 to get \frac{5}{9}. Then add the square of \frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{9}x+\left(\frac{5}{9}\right)^{2}=\frac{4}{3}+\left(\frac{5}{9}\right)^{2}$$

Square \frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{9}x+\frac{25}{81}=\frac{4}{3}+\frac{25}{81}$$

Add \frac{4}{3} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{9}x+\frac{25}{81}=\frac{133}{81}$$

Factor x^{2}+\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{9}\right)^{2}=\frac{133}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{9}\right)^{2}}=\sqrt{\frac{133}{81}}$$

Simplify.

$$x+\frac{5}{9}=\frac{\sqrt{133}}{9} x+\frac{5}{9}=-\frac{\sqrt{133}}{9}$$

Subtract \frac{5}{9} from both sides of the equation.

$$x=\frac{\sqrt{133}-5}{9} x=\frac{-\sqrt{133}-5}{9}$$