Solve 2x^2-5=3x

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

Solve for x

$$x=-1$$

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

Steps Using Factoring By Grouping

Subtract 3x from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-5\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite 2x^{2}-3x-5 as \left(2x^{2}-5x\right)+\left(2x-5\right).

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

Factor out x in 2x^{2}-5x.

$$x\left(2x-5\right)+2x-5$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(x+1\right)$$

To find equation solutions, solve 2x-5=0 and x+1=0.

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

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

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.

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

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

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

Square -3.

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

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-3\right)±\sqrt{9+40}}{2\times 2}$$

Add 9 to 40.

$$x=\frac{-\left(-3\right)±\sqrt{49}}{2\times 2}$$

Take the square root of 49.

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

The opposite of -3 is 3.

$$x=\frac{3±7}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±7}{4}$$

Now solve the equation x=\frac{3±7}{4} when ± is plus. Add 3 to 7.

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

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{3±7}{4} when ± is minus. Subtract 7 from 3.

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

Divide -4 by 4.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 3x from both sides.

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

Add 5 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{5}{2}+\frac{9}{16}$$

Add \frac{5}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{49}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{7}{4} x-\frac{3}{4}=-\frac{7}{4}$$

Add \frac{3}{4} to both sides of the equation.

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