Solve 5x^2=11x-2

$$5 { x }^{ 2 } = 11x-2$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=2$$

Steps Using Factoring By Grouping

Subtract 11x from both sides.

$$5x^{2}-11x=-2$$

Add 2 to both sides.

$$5x^{2}-11x+2=0$$

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

$$a+b=-11 ab=5\times 2=10$$

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 10.

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

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite 5x^{2}-11x+2 as \left(5x^{2}-10x\right)+\left(-x+2\right).

$$\left(5x^{2}-10x\right)+\left(-x+2\right)$$

Factor out 5x in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$5x^{2}-11x=-2$$

Add 2 to both sides.

$$5x^{2}-11x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -11 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 5\times 2}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-11\right)±\sqrt{121-20\times 2}}{2\times 5}$$

Multiply -20 times 2.

$$x=\frac{-\left(-11\right)±\sqrt{121-40}}{2\times 5}$$

Add 121 to -40.

$$x=\frac{-\left(-11\right)±\sqrt{81}}{2\times 5}$$

Take the square root of 81.

$$x=\frac{-\left(-11\right)±9}{2\times 5}$$

The opposite of -11 is 11.

$$x=\frac{11±9}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{11±9}{10}$$

Now solve the equation x=\frac{11±9}{10} when ± is plus. Add 11 to 9.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{11±9}{10} when ± is minus. Subtract 9 from 11.

$$x=\frac{2}{10}$$

Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract 11x from both sides.

$$5x^{2}-11x=-2$$

Divide both sides by 5.

$$\frac{5x^{2}-11x}{5}=-\frac{2}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{11}{5}x=-\frac{2}{5}$$

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Add -\frac{2}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

Factor x^{2}-\frac{11}{5}x+\frac{121}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{9}{10} x-\frac{11}{10}=-\frac{9}{10}$$

Add \frac{11}{10} to both sides of the equation.

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