Solve 5x^2+6=31x

$$5 x ^ { 2 } + 6 = 31 x$$

Solve for x

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

$$x=6$$

Steps Using Factoring By Grouping

Subtract 31x from both sides.

$$5x^{2}+6-31x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$5x^{2}-31x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=-31 ab=5\times 6=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -31.

$$a=-30 b=-1$$

Rewrite 5x^{2}-31x+6 as \left(5x^{2}-30x\right)+\left(-x+6\right).

$$\left(5x^{2}-30x\right)+\left(-x+6\right)$$

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

$$5x\left(x-6\right)-\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(5x-1\right)$$

To find equation solutions, solve x-6=0 and 5x-1=0.

$$x=6 x=\frac{1}{5}$$

Steps Using the Quadratic Formula

Subtract 31x from both sides.

$$5x^{2}+6-31x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$5x^{2}-31x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -31 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 5\times 6}}{2\times 5}$$

Square -31.

$$x=\frac{-\left(-31\right)±\sqrt{961-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-31\right)±\sqrt{961-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$x=\frac{-\left(-31\right)±\sqrt{961-120}}{2\times 5}$$

Add 961 to -120.

$$x=\frac{-\left(-31\right)±\sqrt{841}}{2\times 5}$$

Take the square root of 841.

$$x=\frac{-\left(-31\right)±29}{2\times 5}$$

The opposite of -31 is 31.

$$x=\frac{31±29}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{31±29}{10}$$

Now solve the equation x=\frac{31±29}{10} when ± is plus. Add 31 to 29.

$$x=\frac{60}{10}$$

Divide 60 by 10.

$$x=6$$

Now solve the equation x=\frac{31±29}{10} when ± is minus. Subtract 29 from 31.

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

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

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

The equation is now solved.

$$x=6 x=\frac{1}{5}$$

Steps for Completing the Square

Subtract 31x from both sides.

$$5x^{2}+6-31x=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-31x=-6$$

Divide both sides by 5.

$$\frac{5x^{2}-31x}{5}=-\frac{6}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{31}{5}x=-\frac{6}{5}$$

Divide -\frac{31}{5}, the coefficient of the x term, by 2 to get -\frac{31}{10}. Then add the square of -\frac{31}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{31}{5}x+\left(-\frac{31}{10}\right)^{2}=-\frac{6}{5}+\left(-\frac{31}{10}\right)^{2}$$

Square -\frac{31}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{31}{5}x+\frac{961}{100}=-\frac{6}{5}+\frac{961}{100}$$

Add -\frac{6}{5} to \frac{961}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{31}{5}x+\frac{961}{100}=\frac{841}{100}$$

Factor x^{2}-\frac{31}{5}x+\frac{961}{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}.

$$\left(x-\frac{31}{10}\right)^{2}=\frac{841}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{31}{10}\right)^{2}}=\sqrt{\frac{841}{100}}$$

Simplify.

$$x-\frac{31}{10}=\frac{29}{10} x-\frac{31}{10}=-\frac{29}{10}$$

Add \frac{31}{10} to both sides of the equation.

$$x=6 x=\frac{1}{5}$$