Solve 2x^2+3=5x+3

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

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=0$$

Steps Using Factoring

Subtract 5x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

Factor out x.

$$x\left(2x-5\right)=0$$

To find equation solutions, solve x=0 and 2x-5=0.

$$x=0 x=\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

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

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

Take the square root of \left(-5\right)^{2}.

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

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±5}{4}$$

Now solve the equation x=\frac{5±5}{4} when ± is plus. Add 5 to 5.

$$x=\frac{10}{4}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{5±5}{4} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{2} x=0$$

Steps for Completing the Square

Subtract 5x from both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 3 to get 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 0 by 2.

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

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=\frac{5}{2} x=0$$