Solve 2x^2=7x+4

$$2 { x }^{ 2 } =7x+4$$

Solve for x

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

$$x=4$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

$$2x^{2}-7x=4$$

Subtract 4 from both sides.

$$2x^{2}-7x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\left(-4\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite 2x^{2}-7x-4 as \left(2x^{2}-8x\right)+\left(x-4\right).

$$\left(2x^{2}-8x\right)+\left(x-4\right)$$

Factor out 2x in 2x^{2}-8x.

$$2x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x+1\right)$$

To find equation solutions, solve x-4=0 and 2x+1=0.

$$x=4 x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Subtract 7x from both sides.

$$2x^{2}-7x=4$$

Subtract 4 from both sides.

$$2x^{2}-7x-4=0$$

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

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

Square -7.

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

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\times 2}$$

Add 49 to 32.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2\times 2}$$

Take the square root of 81.

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

The opposite of -7 is 7.

$$x=\frac{7±9}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±9}{4}$$

Now solve the equation x=\frac{7±9}{4} when ± is plus. Add 7 to 9.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{7±9}{4} when ± is minus. Subtract 9 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.

$$x=4 x=-\frac{1}{2}$$

Steps for Completing the Square

Subtract 7x from both sides.

$$2x^{2}-7x=4$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{4}{2}$$

Divide 4 by 2.

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

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.

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

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

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

Add 2 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{81}{16}$$

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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{9}{4} x-\frac{7}{4}=-\frac{9}{4}$$

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

$$x=4 x=-\frac{1}{2}$$