Solve 10x^2-14=31x

$$10 x ^ { 2 } - 14 = 31 x$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using Factoring By Grouping

Subtract 31x from both sides.

$$10x^{2}-14-31x=0$$

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

$$10x^{2}-31x-14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-14. To find a and b, set up a system to be solved.

$$a+b=-31 ab=10\left(-14\right)=-140$$

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 -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -31.

$$a=-35 b=4$$

Rewrite 10x^{2}-31x-14 as \left(10x^{2}-35x\right)+\left(4x-14\right).

$$\left(10x^{2}-35x\right)+\left(4x-14\right)$$

Factor out 5x in the first and 2 in the second group.

$$5x\left(2x-7\right)+2\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(5x+2\right)$$

To find equation solutions, solve 2x-7=0 and 5x+2=0.

$$x=\frac{7}{2} x=-\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 31x from both sides.

$$10x^{2}-14-31x=0$$

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.

$$10x^{2}-31x-14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -31 for b, and -14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 10\left(-14\right)}}{2\times 10}$$

Square -31.

$$x=\frac{-\left(-31\right)±\sqrt{961-4\times 10\left(-14\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-31\right)±\sqrt{961-40\left(-14\right)}}{2\times 10}$$

Multiply -40 times -14.

$$x=\frac{-\left(-31\right)±\sqrt{961+560}}{2\times 10}$$

Add 961 to 560.

$$x=\frac{-\left(-31\right)±\sqrt{1521}}{2\times 10}$$

Take the square root of 1521.

$$x=\frac{-\left(-31\right)±39}{2\times 10}$$

The opposite of -31 is 31.

$$x=\frac{31±39}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{31±39}{20}$$

Now solve the equation x=\frac{31±39}{20} when ± is plus. Add 31 to 39.

$$x=\frac{70}{20}$$

Reduce the fraction \frac{70}{20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{31±39}{20} when ± is minus. Subtract 39 from 31.

$$x=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=\frac{7}{2} x=-\frac{2}{5}$$

Steps for Completing the Square

Subtract 31x from both sides.

$$10x^{2}-14-31x=0$$

Add 14 to both sides. Anything plus zero gives itself.

$$10x^{2}-31x=14$$

Divide both sides by 10.

$$\frac{10x^{2}-31x}{10}=\frac{14}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{31}{10}x=\frac{14}{10}$$

Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{31}{10}x=\frac{7}{5}$$

Divide -\frac{31}{10}, the coefficient of the x term, by 2 to get -\frac{31}{20}. Then add the square of -\frac{31}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{31}{10}x+\left(-\frac{31}{20}\right)^{2}=\frac{7}{5}+\left(-\frac{31}{20}\right)^{2}$$

Square -\frac{31}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{31}{10}x+\frac{961}{400}=\frac{7}{5}+\frac{961}{400}$$

Add \frac{7}{5} to \frac{961}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{31}{10}x+\frac{961}{400}=\frac{1521}{400}$$

Factor x^{2}-\frac{31}{10}x+\frac{961}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{31}{20}\right)^{2}=\frac{1521}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{31}{20}\right)^{2}}=\sqrt{\frac{1521}{400}}$$

Simplify.

$$x-\frac{31}{20}=\frac{39}{20} x-\frac{31}{20}=-\frac{39}{20}$$

Add \frac{31}{20} to both sides of the equation.

$$x=\frac{7}{2} x=-\frac{2}{5}$$