Solve 4x^2=7x+2

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

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 7x from both sides.

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

Subtract 2 from both sides.

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

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

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

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

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

Factor out 4x in 4x^{2}-8x.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 7x from both sides.

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

Subtract 2 from both sides.

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

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

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

Square -7.

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

Multiply -4 times 4.

$$x=\frac{-\left(-7\right)±\sqrt{49-16\left(-2\right)}}{2\times 4}$$

Multiply -16 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\times 4}$$

Add 49 to 32.

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

Take the square root of 81.

$$x=\frac{-\left(-7\right)±9}{2\times 4}$$

The opposite of -7 is 7.

$$x=\frac{7±9}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{7±9}{8}$$

Now solve the equation x=\frac{7±9}{8} when ± is plus. Add 7 to 9.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{7±9}{8} when ± is minus. Subtract 9 from 7.

$$x=-\frac{2}{8}$$

Reduce the fraction \frac{-2}{8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 7x from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

$$x^{2}-\frac{7}{4}x=\frac{1}{2}$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=\frac{1}{2}+\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{1}{2}+\frac{49}{64}$$

Add \frac{1}{2} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{81}{64}$$

Factor x^{2}-\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{81}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{81}{64}}$$

Simplify.

$$x-\frac{7}{8}=\frac{9}{8} x-\frac{7}{8}=-\frac{9}{8}$$

Add \frac{7}{8} to both sides of the equation.

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