Solve 5x^2=3x-7

$$5 x ^ { 2 } = 3 x - 7$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{131}i}{10}\approx 0.3+1.144552314i$$

$$x=\frac{-\sqrt{131}i+3}{10}\approx 0.3-1.144552314i$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$5x^{2}-3x=-7$$

Add 7 to both sides.

$$5x^{2}-3x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -3 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 5\times 7}}{2\times 5}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 5\times 7}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-3\right)±\sqrt{9-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$x=\frac{-\left(-3\right)±\sqrt{9-140}}{2\times 5}$$

Add 9 to -140.

$$x=\frac{-\left(-3\right)±\sqrt{-131}}{2\times 5}$$

Take the square root of -131.

$$x=\frac{-\left(-3\right)±\sqrt{131}i}{2\times 5}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{131}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{3±\sqrt{131}i}{10}$$

Now solve the equation x=\frac{3±\sqrt{131}i}{10} when ± is plus. Add 3 to i\sqrt{131}.

$$x=\frac{3+\sqrt{131}i}{10}$$

Now solve the equation x=\frac{3±\sqrt{131}i}{10} when ± is minus. Subtract i\sqrt{131} from 3.

$$x=\frac{-\sqrt{131}i+3}{10}$$

The equation is now solved.

$$x=\frac{3+\sqrt{131}i}{10} x=\frac{-\sqrt{131}i+3}{10}$$

Steps for Completing the Square

Subtract 3x from both sides.

$$5x^{2}-3x=-7$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

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.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=-\frac{7}{5}+\left(-\frac{3}{10}\right)^{2}$$

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

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=-\frac{7}{5}+\frac{9}{100}$$

Add -\frac{7}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=-\frac{131}{100}$$

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}.

$$\left(x-\frac{3}{10}\right)^{2}=-\frac{131}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{-\frac{131}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{\sqrt{131}i}{10} x-\frac{3}{10}=-\frac{\sqrt{131}i}{10}$$

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

$$x=\frac{3+\sqrt{131}i}{10} x=\frac{-\sqrt{131}i+3}{10}$$