Solve 0=-4/3x^2+19/6x+5/2

$$0 = - \frac { 4 } { 3 } x ^ { 2 } + \frac { 19 } { 6 } x + \frac { 5 } { 2 }$$

Solve for x

$$x=-\frac{5}{8}=-0.625$$

$$x=3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-\frac{4}{3}x^{2}+\frac{19}{6}x+\frac{5}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{4}{3} for a, \frac{19}{6} for b, and \frac{5}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{19}{6}±\sqrt{\left(\frac{19}{6}\right)^{2}-4\left(-\frac{4}{3}\right)\times \frac{5}{2}}}{2\left(-\frac{4}{3}\right)}$$

Square \frac{19}{6} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{19}{6}±\sqrt{\frac{361}{36}-4\left(-\frac{4}{3}\right)\times \frac{5}{2}}}{2\left(-\frac{4}{3}\right)}$$

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

$$x=\frac{-\frac{19}{6}±\sqrt{\frac{361}{36}+\frac{16}{3}\times \frac{5}{2}}}{2\left(-\frac{4}{3}\right)}$$

Multiply \frac{16}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{19}{6}±\sqrt{\frac{361}{36}+\frac{40}{3}}}{2\left(-\frac{4}{3}\right)}$$

Add \frac{361}{36} to \frac{40}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{19}{6}±\sqrt{\frac{841}{36}}}{2\left(-\frac{4}{3}\right)}$$

Take the square root of \frac{841}{36}.

$$x=\frac{-\frac{19}{6}±\frac{29}{6}}{2\left(-\frac{4}{3}\right)}$$

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

$$x=\frac{-\frac{19}{6}±\frac{29}{6}}{-\frac{8}{3}}$$

Now solve the equation x=\frac{-\frac{19}{6}±\frac{29}{6}}{-\frac{8}{3}} when ± is plus. Add -\frac{19}{6} to \frac{29}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{5}{3}}{-\frac{8}{3}}$$

Divide \frac{5}{3} by -\frac{8}{3} by multiplying \frac{5}{3} by the reciprocal of -\frac{8}{3}.

$$x=-\frac{5}{8}$$

Now solve the equation x=\frac{-\frac{19}{6}±\frac{29}{6}}{-\frac{8}{3}} when ± is minus. Subtract \frac{29}{6} from -\frac{19}{6} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{8}{-\frac{8}{3}}$$

Divide -8 by -\frac{8}{3} by multiplying -8 by the reciprocal of -\frac{8}{3}.

$$x=3$$

The equation is now solved.

$$x=-\frac{5}{8} x=3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-\frac{4}{3}x^{2}+\frac{19}{6}x+\frac{5}{2}=0$$

Subtract \frac{5}{2} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{4}{3}x^{2}+\frac{19}{6}x=-\frac{5}{2}$$

Divide both sides of the equation by -\frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{4}{3}x^{2}+\frac{19}{6}x}{-\frac{4}{3}}=-\frac{\frac{5}{2}}{-\frac{4}{3}}$$

Dividing by -\frac{4}{3} undoes the multiplication by -\frac{4}{3}.

$$x^{2}+\frac{\frac{19}{6}}{-\frac{4}{3}}x=-\frac{\frac{5}{2}}{-\frac{4}{3}}$$

Divide \frac{19}{6} by -\frac{4}{3} by multiplying \frac{19}{6} by the reciprocal of -\frac{4}{3}.

$$x^{2}-\frac{19}{8}x=-\frac{\frac{5}{2}}{-\frac{4}{3}}$$

Divide -\frac{5}{2} by -\frac{4}{3} by multiplying -\frac{5}{2} by the reciprocal of -\frac{4}{3}.

$$x^{2}-\frac{19}{8}x=\frac{15}{8}$$

Divide -\frac{19}{8}, the coefficient of the x term, by 2 to get -\frac{19}{16}. Then add the square of -\frac{19}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{8}x+\left(-\frac{19}{16}\right)^{2}=\frac{15}{8}+\left(-\frac{19}{16}\right)^{2}$$

Square -\frac{19}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{8}x+\frac{361}{256}=\frac{15}{8}+\frac{361}{256}$$

Add \frac{15}{8} to \frac{361}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{19}{8}x+\frac{361}{256}=\frac{841}{256}$$

Factor x^{2}-\frac{19}{8}x+\frac{361}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{16}\right)^{2}=\frac{841}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{16}\right)^{2}}=\sqrt{\frac{841}{256}}$$

Simplify.

$$x-\frac{19}{16}=\frac{29}{16} x-\frac{19}{16}=-\frac{29}{16}$$

Add \frac{19}{16} to both sides of the equation.

$$x=3 x=-\frac{5}{8}$$