Solve (x)=2/3x(2x+9)-5x+1

$$( x ) = \frac { 2 } { 3 } x ( 2 x + 9 ) - 5 x + 1$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{3}i}{2}\approx -0-0.866025404i$$

$$x=\frac{\sqrt{3}i}{2}\approx 0.866025404i$$

Steps by Finding Square Root

Use the distributive property to multiply \frac{2}{3}x by 2x+9.

$$x=\frac{2}{3}x\times 2x+\frac{2}{3}x\times 9-5x+1$$

Multiply x and x to get x^{2}.

$$x=\frac{2}{3}x^{2}\times 2+\frac{2}{3}x\times 9-5x+1$$

Express \frac{2}{3}\times 2 as a single fraction.

$$x=\frac{2\times 2}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Multiply 2 and 2 to get 4.

$$x=\frac{4}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Express \frac{2}{3}\times 9 as a single fraction.

$$x=\frac{4}{3}x^{2}+\frac{2\times 9}{3}x-5x+1$$

Multiply 2 and 9 to get 18.

$$x=\frac{4}{3}x^{2}+\frac{18}{3}x-5x+1$$

Divide 18 by 3 to get 6.

$$x=\frac{4}{3}x^{2}+6x-5x+1$$

Combine 6x and -5x to get x.

$$x=\frac{4}{3}x^{2}+x+1$$

Subtract \frac{4}{3}x^{2} from both sides.

$$x-\frac{4}{3}x^{2}=x+1$$

Subtract x from both sides.

$$x-\frac{4}{3}x^{2}-x=1$$

Combine x and -x to get 0.

$$-\frac{4}{3}x^{2}=1$$

Multiply both sides by -\frac{3}{4}, the reciprocal of -\frac{4}{3}.

$$x^{2}=1\left(-\frac{3}{4}\right)$$

Multiply 1 and -\frac{3}{4} to get -\frac{3}{4}.

$$x^{2}=-\frac{3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{3}i}{2} x=-\frac{\sqrt{3}i}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{2}{3}x by 2x+9.

$$x=\frac{2}{3}x\times 2x+\frac{2}{3}x\times 9-5x+1$$

Multiply x and x to get x^{2}.

$$x=\frac{2}{3}x^{2}\times 2+\frac{2}{3}x\times 9-5x+1$$

Express \frac{2}{3}\times 2 as a single fraction.

$$x=\frac{2\times 2}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Multiply 2 and 2 to get 4.

$$x=\frac{4}{3}x^{2}+\frac{2}{3}x\times 9-5x+1$$

Express \frac{2}{3}\times 9 as a single fraction.

$$x=\frac{4}{3}x^{2}+\frac{2\times 9}{3}x-5x+1$$

Multiply 2 and 9 to get 18.

$$x=\frac{4}{3}x^{2}+\frac{18}{3}x-5x+1$$

Divide 18 by 3 to get 6.

$$x=\frac{4}{3}x^{2}+6x-5x+1$$

Combine 6x and -5x to get x.

$$x=\frac{4}{3}x^{2}+x+1$$

Subtract \frac{4}{3}x^{2} from both sides.

$$x-\frac{4}{3}x^{2}=x+1$$

Subtract x from both sides.

$$x-\frac{4}{3}x^{2}-x=1$$

Combine x and -x to get 0.

$$-\frac{4}{3}x^{2}=1$$

Subtract 1 from both sides.

$$-\frac{4}{3}x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{4}{3} for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{4}{3}\right)\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{4}{3}\right)\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Multiply -4 times -\frac{4}{3}.

$$x=\frac{0±\sqrt{\frac{16}{3}\left(-1\right)}}{2\left(-\frac{4}{3}\right)}$$

Multiply \frac{16}{3} times -1.

$$x=\frac{0±\sqrt{-\frac{16}{3}}}{2\left(-\frac{4}{3}\right)}$$

Take the square root of -\frac{16}{3}.

$$x=\frac{0±\frac{4\sqrt{3}i}{3}}{2\left(-\frac{4}{3}\right)}$$

Multiply 2 times -\frac{4}{3}.

$$x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}}$$

Now solve the equation x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}} when ± is plus.

$$x=-\frac{\sqrt{3}i}{2}$$

Now solve the equation x=\frac{0±\frac{4\sqrt{3}i}{3}}{-\frac{8}{3}} when ± is minus.

$$x=\frac{\sqrt{3}i}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{3}i}{2} x=\frac{\sqrt{3}i}{2}$$