Solve -2/5x+7/5=-3/2x^2+1/2x+2

$$- \frac { 2 } { 5 } x + \frac { 7 } { 5 } = - \frac { 3 } { 2 } x ^ { 2 } + \frac { 1 } { 2 } x + 2$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x=1$$

Steps Using the Quadratic Formula

Add \frac{3}{2}x^{2} to both sides.

$$-\frac{2}{5}x+\frac{7}{5}+\frac{3}{2}x^{2}=\frac{1}{2}x+2$$

Subtract \frac{1}{2}x from both sides.

$$-\frac{2}{5}x+\frac{7}{5}+\frac{3}{2}x^{2}-\frac{1}{2}x=2$$

Combine -\frac{2}{5}x and -\frac{1}{2}x to get -\frac{9}{10}x.

$$-\frac{9}{10}x+\frac{7}{5}+\frac{3}{2}x^{2}=2$$

Subtract 2 from both sides.

$$-\frac{9}{10}x+\frac{7}{5}+\frac{3}{2}x^{2}-2=0$$

Subtract 2 from \frac{7}{5} to get -\frac{3}{5}.

$$-\frac{9}{10}x-\frac{3}{5}+\frac{3}{2}x^{2}=0$$

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.

$$\frac{3}{2}x^{2}-\frac{9}{10}x-\frac{3}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -\frac{9}{10} for b, and -\frac{3}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{9}{10}\right)±\sqrt{\left(-\frac{9}{10}\right)^{2}-4\times \frac{3}{2}\left(-\frac{3}{5}\right)}}{2\times \frac{3}{2}}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{9}{10}\right)±\sqrt{\frac{81}{100}-4\times \frac{3}{2}\left(-\frac{3}{5}\right)}}{2\times \frac{3}{2}}$$

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

$$x=\frac{-\left(-\frac{9}{10}\right)±\sqrt{\frac{81}{100}-6\left(-\frac{3}{5}\right)}}{2\times \frac{3}{2}}$$

Multiply -6 times -\frac{3}{5}.

$$x=\frac{-\left(-\frac{9}{10}\right)±\sqrt{\frac{81}{100}+\frac{18}{5}}}{2\times \frac{3}{2}}$$

Add \frac{81}{100} to \frac{18}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{9}{10}\right)±\sqrt{\frac{441}{100}}}{2\times \frac{3}{2}}$$

Take the square root of \frac{441}{100}.

$$x=\frac{-\left(-\frac{9}{10}\right)±\frac{21}{10}}{2\times \frac{3}{2}}$$

The opposite of -\frac{9}{10} is \frac{9}{10}.

$$x=\frac{\frac{9}{10}±\frac{21}{10}}{2\times \frac{3}{2}}$$

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

$$x=\frac{\frac{9}{10}±\frac{21}{10}}{3}$$

Now solve the equation x=\frac{\frac{9}{10}±\frac{21}{10}}{3} when ± is plus. Add \frac{9}{10} to \frac{21}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide 3 by 3.

$$x=1$$

Now solve the equation x=\frac{\frac{9}{10}±\frac{21}{10}}{3} when ± is minus. Subtract \frac{21}{10} from \frac{9}{10} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{6}{5}}{3}$$

Divide -\frac{6}{5} by 3.

$$x=-\frac{2}{5}$$

The equation is now solved.

$$x=1 x=-\frac{2}{5}$$

Steps for Completing the Square

Add \frac{3}{2}x^{2} to both sides.

$$-\frac{2}{5}x+\frac{7}{5}+\frac{3}{2}x^{2}=\frac{1}{2}x+2$$

Subtract \frac{1}{2}x from both sides.

$$-\frac{2}{5}x+\frac{7}{5}+\frac{3}{2}x^{2}-\frac{1}{2}x=2$$

Combine -\frac{2}{5}x and -\frac{1}{2}x to get -\frac{9}{10}x.

$$-\frac{9}{10}x+\frac{7}{5}+\frac{3}{2}x^{2}=2$$

Subtract \frac{7}{5} from both sides.

$$-\frac{9}{10}x+\frac{3}{2}x^{2}=2-\frac{7}{5}$$

Subtract \frac{7}{5} from 2 to get \frac{3}{5}.

$$-\frac{9}{10}x+\frac{3}{2}x^{2}=\frac{3}{5}$$

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.

$$\frac{3}{2}x^{2}-\frac{9}{10}x=\frac{3}{5}$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2}x^{2}-\frac{9}{10}x}{\frac{3}{2}}=\frac{\frac{3}{5}}{\frac{3}{2}}$$

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

$$x^{2}+\left(-\frac{\frac{9}{10}}{\frac{3}{2}}\right)x=\frac{\frac{3}{5}}{\frac{3}{2}}$$

Divide -\frac{9}{10} by \frac{3}{2} by multiplying -\frac{9}{10} by the reciprocal of \frac{3}{2}.

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

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

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

Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\frac{2}{5}+\left(-\frac{3}{10}\right)^{2}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{2}{5}+\frac{9}{100}$$

Add \frac{2}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{49}{100}$$

Factor x^{2}-\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{7}{10} x-\frac{3}{10}=-\frac{7}{10}$$

Add \frac{3}{10} to both sides of the equation.

$$x=1 x=-\frac{2}{5}$$