Solve 3x^2=22x-7

$$3 x ^ { 2 } = 22 x - 7$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=7$$

Steps Using Factoring By Grouping

Subtract 22x from both sides.

$$3x^{2}-22x=-7$$

Add 7 to both sides.

$$3x^{2}-22x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-22 ab=3\times 7=21$$

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 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -22.

$$a=-21 b=-1$$

Rewrite 3x^{2}-22x+7 as \left(3x^{2}-21x\right)+\left(-x+7\right).

$$\left(3x^{2}-21x\right)+\left(-x+7\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-7\right)-\left(x-7\right)$$

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

$$\left(x-7\right)\left(3x-1\right)$$

To find equation solutions, solve x-7=0 and 3x-1=0.

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

Steps Using the Quadratic Formula

Subtract 22x from both sides.

$$3x^{2}-22x=-7$$

Add 7 to both sides.

$$3x^{2}-22x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -22 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-22\right)±\sqrt{484-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$x=\frac{-\left(-22\right)±\sqrt{484-84}}{2\times 3}$$

Add 484 to -84.

$$x=\frac{-\left(-22\right)±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-\left(-22\right)±20}{2\times 3}$$

The opposite of -22 is 22.

$$x=\frac{22±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{22±20}{6}$$

Now solve the equation x=\frac{22±20}{6} when ± is plus. Add 22 to 20.

$$x=\frac{42}{6}$$

Divide 42 by 6.

$$x=7$$

Now solve the equation x=\frac{22±20}{6} when ± is minus. Subtract 20 from 22.

$$x=\frac{2}{6}$$

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

$$x=\frac{1}{3}$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 22x from both sides.

$$3x^{2}-22x=-7$$

Divide both sides by 3.

$$\frac{3x^{2}-22x}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{22}{3}x=-\frac{7}{3}$$

Divide -\frac{22}{3}, the coefficient of the x term, by 2 to get -\frac{11}{3}. Then add the square of -\frac{11}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{22}{3}x+\left(-\frac{11}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{11}{3}\right)^{2}$$

Square -\frac{11}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=-\frac{7}{3}+\frac{121}{9}$$

Add -\frac{7}{3} to \frac{121}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{22}{3}x+\frac{121}{9}=\frac{100}{9}$$

Factor x^{2}-\frac{22}{3}x+\frac{121}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x-\frac{11}{3}=\frac{10}{3} x-\frac{11}{3}=-\frac{10}{3}$$

Add \frac{11}{3} to both sides of the equation.

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