Solve 2x^2=7x-3

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

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

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

Add 3 to both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 2x^{2}-7x+3 as \left(2x^{2}-6x\right)+\left(-x+3\right).

$$\left(2x^{2}-6x\right)+\left(-x+3\right)$$

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x-1\right)$$

To find equation solutions, solve x-3=0 and 2x-1=0.

$$x=3 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

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

Add 3 to both sides.

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

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

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

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 3}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\times 3}}{2\times 2}$$

Multiply -8 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\times 2}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\times 2}$$

Take the square root of 25.

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

The opposite of -7 is 7.

$$x=\frac{7±5}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±5}{4}$$

Now solve the equation x=\frac{7±5}{4} when ± is plus. Add 7 to 5.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{7±5}{4} when ± is minus. Subtract 5 from 7.

$$x=\frac{2}{4}$$

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

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

The equation is now solved.

$$x=3 x=\frac{1}{2}$$

Steps for Completing the Square

Subtract 7x from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{5}{4} x-\frac{7}{4}=-\frac{5}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=3 x=\frac{1}{2}$$