Solve 4x^2=11x-6

$$4 x ^ { 2 } = 11 x - 6$$

Solve for x

$$x=2$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Subtract 11x from both sides.

$$4x^{2}-11x=-6$$

Add 6 to both sides.

$$4x^{2}-11x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-11 ab=4\times 6=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

Rewrite 4x^{2}-11x+6 as \left(4x^{2}-8x\right)+\left(-3x+6\right).

$$\left(4x^{2}-8x\right)+\left(-3x+6\right)$$

Factor out 4x in the first and -3 in the second group.

$$4x\left(x-2\right)-3\left(x-2\right)$$

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

$$\left(x-2\right)\left(4x-3\right)$$

To find equation solutions, solve x-2=0 and 4x-3=0.

$$x=2 x=\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$4x^{2}-11x=-6$$

Add 6 to both sides.

$$4x^{2}-11x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -11 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 4}$$

Add 121 to -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2\times 4}$$

Take the square root of 25.

$$x=\frac{-\left(-11\right)±5}{2\times 4}$$

The opposite of -11 is 11.

$$x=\frac{11±5}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{11±5}{8}$$

Now solve the equation x=\frac{11±5}{8} when ± is plus. Add 11 to 5.

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

Divide 16 by 8.

$$x=2$$

Now solve the equation x=\frac{11±5}{8} when ± is minus. Subtract 5 from 11.

$$x=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=2 x=\frac{3}{4}$$

Steps for Completing the Square

Subtract 11x from both sides.

$$4x^{2}-11x=-6$$

Divide both sides by 4.

$$\frac{4x^{2}-11x}{4}=-\frac{6}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{11}{4}x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{4}x=-\frac{3}{2}$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=-\frac{3}{2}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=-\frac{3}{2}+\frac{121}{64}$$

Add -\frac{3}{2} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{25}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{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}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{5}{8} x-\frac{11}{8}=-\frac{5}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=2 x=\frac{3}{4}$$