Solve (6x-3)*(5x+2)=30x

$$( 6 x - 3 ) \cdot ( 5 x + 2 ) = 30 x$$

Solve for x

$$x = \frac{\sqrt{201} + 11}{20} \approx 1.258872344$$

$$x=\frac{11-\sqrt{201}}{20}\approx -0.158872344$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6x-3 by 5x+2 and combine like terms.

$$30x^{2}-3x-6=30x$$

Subtract 30x from both sides.

$$30x^{2}-3x-6-30x=0$$

Combine -3x and -30x to get -33x.

$$30x^{2}-33x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 30 for a, -33 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 30\left(-6\right)}}{2\times 30}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\times 30\left(-6\right)}}{2\times 30}$$

Multiply -4 times 30.

$$x=\frac{-\left(-33\right)±\sqrt{1089-120\left(-6\right)}}{2\times 30}$$

Multiply -120 times -6.

$$x=\frac{-\left(-33\right)±\sqrt{1089+720}}{2\times 30}$$

Add 1089 to 720.

$$x=\frac{-\left(-33\right)±\sqrt{1809}}{2\times 30}$$

Take the square root of 1809.

$$x=\frac{-\left(-33\right)±3\sqrt{201}}{2\times 30}$$

The opposite of -33 is 33.

$$x=\frac{33±3\sqrt{201}}{2\times 30}$$

Multiply 2 times 30.

$$x=\frac{33±3\sqrt{201}}{60}$$

Now solve the equation x=\frac{33±3\sqrt{201}}{60} when ± is plus. Add 33 to 3\sqrt{201}.

$$x=\frac{3\sqrt{201}+33}{60}$$

Divide 33+3\sqrt{201} by 60.

$$x=\frac{\sqrt{201}+11}{20}$$

Now solve the equation x=\frac{33±3\sqrt{201}}{60} when ± is minus. Subtract 3\sqrt{201} from 33.

$$x=\frac{33-3\sqrt{201}}{60}$$

Divide 33-3\sqrt{201} by 60.

$$x=\frac{11-\sqrt{201}}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{201}+11}{20} x=\frac{11-\sqrt{201}}{20}$$

Steps for Completing the Square

Use the distributive property to multiply 6x-3 by 5x+2 and combine like terms.

$$30x^{2}-3x-6=30x$$

Subtract 30x from both sides.

$$30x^{2}-3x-6-30x=0$$

Combine -3x and -30x to get -33x.

$$30x^{2}-33x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$30x^{2}-33x=6$$

Divide both sides by 30.

$$\frac{30x^{2}-33x}{30}=\frac{6}{30}$$

Dividing by 30 undoes the multiplication by 30.

$$x^{2}+\left(-\frac{33}{30}\right)x=\frac{6}{30}$$

Reduce the fraction \frac{-33}{30} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{11}{10}x=\frac{6}{30}$$

Reduce the fraction \frac{6}{30} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{11}{10}x=\frac{1}{5}$$

Divide -\frac{11}{10}, the coefficient of the x term, by 2 to get -\frac{11}{20}. Then add the square of -\frac{11}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{10}x+\left(-\frac{11}{20}\right)^{2}=\frac{1}{5}+\left(-\frac{11}{20}\right)^{2}$$

Square -\frac{11}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{10}x+\frac{121}{400}=\frac{1}{5}+\frac{121}{400}$$

Add \frac{1}{5} to \frac{121}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{10}x+\frac{121}{400}=\frac{201}{400}$$

Factor x^{2}-\frac{11}{10}x+\frac{121}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{20}\right)^{2}=\frac{201}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{20}\right)^{2}}=\sqrt{\frac{201}{400}}$$

Simplify.

$$x-\frac{11}{20}=\frac{\sqrt{201}}{20} x-\frac{11}{20}=-\frac{\sqrt{201}}{20}$$

Add \frac{11}{20} to both sides of the equation.

$$x=\frac{\sqrt{201}+11}{20} x=\frac{11-\sqrt{201}}{20}$$