Solve 2/3x-1/4+2=-2/9x^2-5/9x+frac{4}{3}

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

Solve for x

$$x=\frac{\sqrt{91}-11}{4}\approx -0.365151996$$

$$x=\frac{-\sqrt{91}-11}{4}\approx -5.134848004$$

Steps Using the Quadratic Formula

Add -\frac{1}{4} and 2 to get \frac{7}{4}.

$$\frac{2}{3}x+\frac{7}{4}=-\frac{2}{9}x^{2}-\frac{5}{9}x+\frac{4}{3}$$

Add \frac{2}{9}x^{2} to both sides.

$$\frac{2}{3}x+\frac{7}{4}+\frac{2}{9}x^{2}=-\frac{5}{9}x+\frac{4}{3}$$

Add \frac{5}{9}x to both sides.

$$\frac{2}{3}x+\frac{7}{4}+\frac{2}{9}x^{2}+\frac{5}{9}x=\frac{4}{3}$$

Combine \frac{2}{3}x and \frac{5}{9}x to get \frac{11}{9}x.

$$\frac{11}{9}x+\frac{7}{4}+\frac{2}{9}x^{2}=\frac{4}{3}$$

Subtract \frac{4}{3} from both sides.

$$\frac{11}{9}x+\frac{7}{4}+\frac{2}{9}x^{2}-\frac{4}{3}=0$$

Subtract \frac{4}{3} from \frac{7}{4} to get \frac{5}{12}.

$$\frac{11}{9}x+\frac{5}{12}+\frac{2}{9}x^{2}=0$$

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.

$$\frac{2}{9}x^{2}+\frac{11}{9}x+\frac{5}{12}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{2}{9} for a, \frac{11}{9} for b, and \frac{5}{12} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{11}{9}±\sqrt{\left(\frac{11}{9}\right)^{2}-4\times \frac{2}{9}\times \frac{5}{12}}}{2\times \frac{2}{9}}$$

Square \frac{11}{9} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{11}{9}±\sqrt{\frac{121}{81}-4\times \frac{2}{9}\times \frac{5}{12}}}{2\times \frac{2}{9}}$$

Multiply -4 times \frac{2}{9}.

$$x=\frac{-\frac{11}{9}±\sqrt{\frac{121}{81}-\frac{8}{9}\times \frac{5}{12}}}{2\times \frac{2}{9}}$$

Multiply -\frac{8}{9} times \frac{5}{12} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{11}{9}±\sqrt{\frac{121}{81}-\frac{10}{27}}}{2\times \frac{2}{9}}$$

Add \frac{121}{81} to -\frac{10}{27} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{11}{9}±\sqrt{\frac{91}{81}}}{2\times \frac{2}{9}}$$

Take the square root of \frac{91}{81}.

$$x=\frac{-\frac{11}{9}±\frac{\sqrt{91}}{9}}{2\times \frac{2}{9}}$$

Multiply 2 times \frac{2}{9}.

$$x=\frac{-\frac{11}{9}±\frac{\sqrt{91}}{9}}{\frac{4}{9}}$$

Now solve the equation x=\frac{-\frac{11}{9}±\frac{\sqrt{91}}{9}}{\frac{4}{9}} when ± is plus. Add -\frac{11}{9} to \frac{\sqrt{91}}{9}.

$$x=\frac{\sqrt{91}-11}{\frac{4}{9}\times 9}$$

Divide \frac{-11+\sqrt{91}}{9} by \frac{4}{9} by multiplying \frac{-11+\sqrt{91}}{9} by the reciprocal of \frac{4}{9}.

$$x=\frac{\sqrt{91}-11}{4}$$

Now solve the equation x=\frac{-\frac{11}{9}±\frac{\sqrt{91}}{9}}{\frac{4}{9}} when ± is minus. Subtract \frac{\sqrt{91}}{9} from -\frac{11}{9}.

$$x=\frac{-\sqrt{91}-11}{\frac{4}{9}\times 9}$$

Divide \frac{-11-\sqrt{91}}{9} by \frac{4}{9} by multiplying \frac{-11-\sqrt{91}}{9} by the reciprocal of \frac{4}{9}.

$$x=\frac{-\sqrt{91}-11}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{91}-11}{4} x=\frac{-\sqrt{91}-11}{4}$$

Steps for Completing the Square

Add -\frac{1}{4} and 2 to get \frac{7}{4}.

$$\frac{2}{3}x+\frac{7}{4}=-\frac{2}{9}x^{2}-\frac{5}{9}x+\frac{4}{3}$$

Add \frac{2}{9}x^{2} to both sides.

$$\frac{2}{3}x+\frac{7}{4}+\frac{2}{9}x^{2}=-\frac{5}{9}x+\frac{4}{3}$$

Add \frac{5}{9}x to both sides.

$$\frac{2}{3}x+\frac{7}{4}+\frac{2}{9}x^{2}+\frac{5}{9}x=\frac{4}{3}$$

Combine \frac{2}{3}x and \frac{5}{9}x to get \frac{11}{9}x.

$$\frac{11}{9}x+\frac{7}{4}+\frac{2}{9}x^{2}=\frac{4}{3}$$

Subtract \frac{7}{4} from both sides.

$$\frac{11}{9}x+\frac{2}{9}x^{2}=\frac{4}{3}-\frac{7}{4}$$

Subtract \frac{7}{4} from \frac{4}{3} to get -\frac{5}{12}.

$$\frac{11}{9}x+\frac{2}{9}x^{2}=-\frac{5}{12}$$

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.

$$\frac{2}{9}x^{2}+\frac{11}{9}x=-\frac{5}{12}$$

Divide both sides of the equation by \frac{2}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{2}{9}x^{2}+\frac{11}{9}x}{\frac{2}{9}}=-\frac{\frac{5}{12}}{\frac{2}{9}}$$

Dividing by \frac{2}{9} undoes the multiplication by \frac{2}{9}.

$$x^{2}+\frac{\frac{11}{9}}{\frac{2}{9}}x=-\frac{\frac{5}{12}}{\frac{2}{9}}$$

Divide \frac{11}{9} by \frac{2}{9} by multiplying \frac{11}{9} by the reciprocal of \frac{2}{9}.

$$x^{2}+\frac{11}{2}x=-\frac{\frac{5}{12}}{\frac{2}{9}}$$

Divide -\frac{5}{12} by \frac{2}{9} by multiplying -\frac{5}{12} by the reciprocal of \frac{2}{9}.

$$x^{2}+\frac{11}{2}x=-\frac{15}{8}$$

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.

$$x^{2}+\frac{11}{2}x+\left(\frac{11}{4}\right)^{2}=-\frac{15}{8}+\left(\frac{11}{4}\right)^{2}$$

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

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=-\frac{15}{8}+\frac{121}{16}$$

Add -\frac{15}{8} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{2}x+\frac{121}{16}=\frac{91}{16}$$

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}.

$$\left(x+\frac{11}{4}\right)^{2}=\frac{91}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{4}\right)^{2}}=\sqrt{\frac{91}{16}}$$

Simplify.

$$x+\frac{11}{4}=\frac{\sqrt{91}}{4} x+\frac{11}{4}=-\frac{\sqrt{91}}{4}$$

Subtract \frac{11}{4} from both sides of the equation.

$$x=\frac{\sqrt{91}-11}{4} x=\frac{-\sqrt{91}-11}{4}$$