Solve 5x^2=23x+42

$$5 x ^ { 2 } = 23 x + 42$$

Solve for x

$$x = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

$$x=6$$

Steps Using Factoring By Grouping

Subtract 23x from both sides.

$$5x^{2}-23x=42$$

Subtract 42 from both sides.

$$5x^{2}-23x-42=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-42. To find a and b, set up a system to be solved.

$$a+b=-23 ab=5\left(-42\right)=-210$$

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 -210.

$$1,-210 2,-105 3,-70 5,-42 6,-35 7,-30 10,-21 14,-15$$

Calculate the sum for each pair.

$$1-210=-209 2-105=-103 3-70=-67 5-42=-37 6-35=-29 7-30=-23 10-21=-11 14-15=-1$$

The solution is the pair that gives sum -23.

$$a=-30 b=7$$

Rewrite 5x^{2}-23x-42 as \left(5x^{2}-30x\right)+\left(7x-42\right).

$$\left(5x^{2}-30x\right)+\left(7x-42\right)$$

Factor out 5x in the first and 7 in the second group.

$$5x\left(x-6\right)+7\left(x-6\right)$$

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

$$\left(x-6\right)\left(5x+7\right)$$

To find equation solutions, solve x-6=0 and 5x+7=0.

$$x=6 x=-\frac{7}{5}$$

Steps Using the Quadratic Formula

Subtract 23x from both sides.

$$5x^{2}-23x=42$$

Subtract 42 from both sides.

$$5x^{2}-23x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -23 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 5\left(-42\right)}}{2\times 5}$$

Square -23.

$$x=\frac{-\left(-23\right)±\sqrt{529-4\times 5\left(-42\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-23\right)±\sqrt{529-20\left(-42\right)}}{2\times 5}$$

Multiply -20 times -42.

$$x=\frac{-\left(-23\right)±\sqrt{529+840}}{2\times 5}$$

Add 529 to 840.

$$x=\frac{-\left(-23\right)±\sqrt{1369}}{2\times 5}$$

Take the square root of 1369.

$$x=\frac{-\left(-23\right)±37}{2\times 5}$$

The opposite of -23 is 23.

$$x=\frac{23±37}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{23±37}{10}$$

Now solve the equation x=\frac{23±37}{10} when ± is plus. Add 23 to 37.

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

Divide 60 by 10.

$$x=6$$

Now solve the equation x=\frac{23±37}{10} when ± is minus. Subtract 37 from 23.

$$x=-\frac{14}{10}$$

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

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

The equation is now solved.

$$x=6 x=-\frac{7}{5}$$

Steps for Completing the Square

Subtract 23x from both sides.

$$5x^{2}-23x=42$$

Divide both sides by 5.

$$\frac{5x^{2}-23x}{5}=\frac{42}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{23}{5}x=\frac{42}{5}$$

Divide -\frac{23}{5}, the coefficient of the x term, by 2 to get -\frac{23}{10}. Then add the square of -\frac{23}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{23}{5}x+\left(-\frac{23}{10}\right)^{2}=\frac{42}{5}+\left(-\frac{23}{10}\right)^{2}$$

Square -\frac{23}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{23}{5}x+\frac{529}{100}=\frac{42}{5}+\frac{529}{100}$$

Add \frac{42}{5} to \frac{529}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{23}{5}x+\frac{529}{100}=\frac{1369}{100}$$

Factor x^{2}-\frac{23}{5}x+\frac{529}{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}.

$$\left(x-\frac{23}{10}\right)^{2}=\frac{1369}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{23}{10}\right)^{2}}=\sqrt{\frac{1369}{100}}$$

Simplify.

$$x-\frac{23}{10}=\frac{37}{10} x-\frac{23}{10}=-\frac{37}{10}$$

Add \frac{23}{10} to both sides of the equation.

$$x=6 x=-\frac{7}{5}$$