Solve 3x^2=5x+3

$$3 x ^ { 2 } = 5 x + 3$$

Solve for x

$$x = \frac{\sqrt{61} + 5}{6} \approx 2.135041613$$

$$x=\frac{5-\sqrt{61}}{6}\approx -0.468374946$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

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

Subtract 3 from both sides.

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

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

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$x=\frac{-\left(-5\right)±\sqrt{25+36}}{2\times 3}$$

Add 25 to 36.

$$x=\frac{-\left(-5\right)±\sqrt{61}}{2\times 3}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{61}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{5±\sqrt{61}}{6}$$

Now solve the equation x=\frac{5±\sqrt{61}}{6} when ± is plus. Add 5 to \sqrt{61}.

$$x=\frac{\sqrt{61}+5}{6}$$

Now solve the equation x=\frac{5±\sqrt{61}}{6} when ± is minus. Subtract \sqrt{61} from 5.

$$x=\frac{5-\sqrt{61}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{61}+5}{6} x=\frac{5-\sqrt{61}}{6}$$

Steps for Completing the Square

Subtract 5x from both sides.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide 3 by 3.

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

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=1+\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=1+\frac{25}{36}$$

Add 1 to \frac{25}{36}.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{61}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{61}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{61}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{\sqrt{61}}{6} x-\frac{5}{6}=-\frac{\sqrt{61}}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{\sqrt{61}+5}{6} x=\frac{5-\sqrt{61}}{6}$$