Solve {x^2+(5x+2/3)^2-2x+5x+2/3-3=0

$$\{ x ^ { 2 } + ( \frac { 5 x + 2 } { 3 } ) ^ { 2 } - 2 x + \frac { 5 x + 2 } { 3 } - 3 = 0$$

Solve for x

$$x=-1$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$3x^{2}+3\times \left(\frac{5x+2}{3}\right)^{2}-6x+5x+2-9=0$$

To raise \frac{5x+2}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3x^{2}+3\times \frac{\left(5x+2\right)^{2}}{3^{2}}-6x+5x+2-9=0$$

Express 3\times \frac{\left(5x+2\right)^{2}}{3^{2}} as a single fraction.

$$3x^{2}+\frac{3\left(5x+2\right)^{2}}{3^{2}}-6x+5x+2-9=0$$

Cancel out 3 in both numerator and denominator.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-6x+5x+2-9=0$$

Combine -6x and 5x to get -x.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-x+2-9=0$$

Subtract 9 from 2 to get -7.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-x-7=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+2\right)^{2}.

$$3x^{2}+\frac{25x^{2}+20x+4}{3}-x-7=0$$

Divide each term of 25x^{2}+20x+4 by 3 to get \frac{25}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}.

$$3x^{2}+\frac{25}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}-x-7=0$$

Combine 3x^{2} and \frac{25}{3}x^{2} to get \frac{34}{3}x^{2}.

$$\frac{34}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}-x-7=0$$

Combine \frac{20}{3}x and -x to get \frac{17}{3}x.

$$\frac{34}{3}x^{2}+\frac{17}{3}x+\frac{4}{3}-7=0$$

Subtract 7 from \frac{4}{3} to get -\frac{17}{3}.

$$\frac{34}{3}x^{2}+\frac{17}{3}x-\frac{17}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{34}{3} for a, \frac{17}{3} for b, and -\frac{17}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{17}{3}±\sqrt{\left(\frac{17}{3}\right)^{2}-4\times \frac{34}{3}\left(-\frac{17}{3}\right)}}{2\times \frac{34}{3}}$$

Square \frac{17}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{17}{3}±\sqrt{\frac{289}{9}-4\times \frac{34}{3}\left(-\frac{17}{3}\right)}}{2\times \frac{34}{3}}$$

Multiply -4 times \frac{34}{3}.

$$x=\frac{-\frac{17}{3}±\sqrt{\frac{289}{9}-\frac{136}{3}\left(-\frac{17}{3}\right)}}{2\times \frac{34}{3}}$$

Multiply -\frac{136}{3} times -\frac{17}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{17}{3}±\sqrt{\frac{289+2312}{9}}}{2\times \frac{34}{3}}$$

Add \frac{289}{9} to \frac{2312}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{17}{3}±\sqrt{289}}{2\times \frac{34}{3}}$$

Take the square root of 289.

$$x=\frac{-\frac{17}{3}±17}{2\times \frac{34}{3}}$$

Multiply 2 times \frac{34}{3}.

$$x=\frac{-\frac{17}{3}±17}{\frac{68}{3}}$$

Now solve the equation x=\frac{-\frac{17}{3}±17}{\frac{68}{3}} when ± is plus. Add -\frac{17}{3} to 17.

$$x=\frac{\frac{34}{3}}{\frac{68}{3}}$$

Divide \frac{34}{3} by \frac{68}{3} by multiplying \frac{34}{3} by the reciprocal of \frac{68}{3}.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-\frac{17}{3}±17}{\frac{68}{3}} when ± is minus. Subtract 17 from -\frac{17}{3}.

$$x=-\frac{\frac{68}{3}}{\frac{68}{3}}$$

Divide -\frac{68}{3} by \frac{68}{3} by multiplying -\frac{68}{3} by the reciprocal of \frac{68}{3}.

$$x=-1$$

The equation is now solved.

$$x=\frac{1}{2} x=-1$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$3x^{2}+3\times \left(\frac{5x+2}{3}\right)^{2}-6x+5x+2-9=0$$

To raise \frac{5x+2}{3} to a power, raise both numerator and denominator to the power and then divide.

$$3x^{2}+3\times \frac{\left(5x+2\right)^{2}}{3^{2}}-6x+5x+2-9=0$$

Express 3\times \frac{\left(5x+2\right)^{2}}{3^{2}} as a single fraction.

$$3x^{2}+\frac{3\left(5x+2\right)^{2}}{3^{2}}-6x+5x+2-9=0$$

Cancel out 3 in both numerator and denominator.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-6x+5x+2-9=0$$

Combine -6x and 5x to get -x.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-x+2-9=0$$

Subtract 9 from 2 to get -7.

$$3x^{2}+\frac{\left(5x+2\right)^{2}}{3}-x-7=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+2\right)^{2}.

$$3x^{2}+\frac{25x^{2}+20x+4}{3}-x-7=0$$

Divide each term of 25x^{2}+20x+4 by 3 to get \frac{25}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}.

$$3x^{2}+\frac{25}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}-x-7=0$$

Combine 3x^{2} and \frac{25}{3}x^{2} to get \frac{34}{3}x^{2}.

$$\frac{34}{3}x^{2}+\frac{20}{3}x+\frac{4}{3}-x-7=0$$

Combine \frac{20}{3}x and -x to get \frac{17}{3}x.

$$\frac{34}{3}x^{2}+\frac{17}{3}x+\frac{4}{3}-7=0$$

Subtract 7 from \frac{4}{3} to get -\frac{17}{3}.

$$\frac{34}{3}x^{2}+\frac{17}{3}x-\frac{17}{3}=0$$

Add \frac{17}{3} to both sides. Anything plus zero gives itself.

$$\frac{34}{3}x^{2}+\frac{17}{3}x=\frac{17}{3}$$

Divide both sides of the equation by \frac{34}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{34}{3}x^{2}+\frac{17}{3}x}{\frac{34}{3}}=\frac{\frac{17}{3}}{\frac{34}{3}}$$

Dividing by \frac{34}{3} undoes the multiplication by \frac{34}{3}.

$$x^{2}+\frac{\frac{17}{3}}{\frac{34}{3}}x=\frac{\frac{17}{3}}{\frac{34}{3}}$$

Divide \frac{17}{3} by \frac{34}{3} by multiplying \frac{17}{3} by the reciprocal of \frac{34}{3}.

$$x^{2}+\frac{1}{2}x=\frac{\frac{17}{3}}{\frac{34}{3}}$$

Divide \frac{17}{3} by \frac{34}{3} by multiplying \frac{17}{3} by the reciprocal of \frac{34}{3}.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Add \frac{1}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

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

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{1}{2} x=-1$$