Solve x+7quad3x^2-18x+2=0

$$x + 7 \quad 3 x ^ { 2 } - 18 x + 2 = 0$$

Solve for x

$$x=\frac{1}{7}\approx 0.142857143$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Multiply 7 and 3 to get 21.

$$x+21x^{2}-18x+2=0$$

Combine x and -18x to get -17x.

$$-17x+21x^{2}+2=0$$

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

$$21x^{2}-17x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 21x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-17 ab=21\times 2=42$$

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 42.

$$-1,-42 -2,-21 -3,-14 -6,-7$$

Calculate the sum for each pair.

$$-1-42=-43 -2-21=-23 -3-14=-17 -6-7=-13$$

The solution is the pair that gives sum -17.

$$a=-14 b=-3$$

Rewrite 21x^{2}-17x+2 as \left(21x^{2}-14x\right)+\left(-3x+2\right).

$$\left(21x^{2}-14x\right)+\left(-3x+2\right)$$

Factor out 7x in the first and -1 in the second group.

$$7x\left(3x-2\right)-\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(7x-1\right)$$

To find equation solutions, solve 3x-2=0 and 7x-1=0.

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

Steps Using the Quadratic Formula

Multiply 7 and 3 to get 21.

$$x+21x^{2}-18x+2=0$$

Combine x and -18x to get -17x.

$$-17x+21x^{2}+2=0$$

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.

$$21x^{2}-17x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -17 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 21\times 2}}{2\times 21}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 21\times 2}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-\left(-17\right)±\sqrt{289-84\times 2}}{2\times 21}$$

Multiply -84 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-168}}{2\times 21}$$

Add 289 to -168.

$$x=\frac{-\left(-17\right)±\sqrt{121}}{2\times 21}$$

Take the square root of 121.

$$x=\frac{-\left(-17\right)±11}{2\times 21}$$

The opposite of -17 is 17.

$$x=\frac{17±11}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{17±11}{42}$$

Now solve the equation x=\frac{17±11}{42} when ± is plus. Add 17 to 11.

$$x=\frac{28}{42}$$

Reduce the fraction \frac{28}{42} to lowest terms by extracting and canceling out 14.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{17±11}{42} when ± is minus. Subtract 11 from 17.

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

Reduce the fraction \frac{6}{42} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply 7 and 3 to get 21.

$$x+21x^{2}-18x+2=0$$

Combine x and -18x to get -17x.

$$-17x+21x^{2}+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$-17x+21x^{2}=-2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$21x^{2}-17x=-2$$

Divide both sides by 21.

$$\frac{21x^{2}-17x}{21}=-\frac{2}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$x^{2}-\frac{17}{21}x=-\frac{2}{21}$$

Divide -\frac{17}{21}, the coefficient of the x term, by 2 to get -\frac{17}{42}. Then add the square of -\frac{17}{42} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{21}x+\left(-\frac{17}{42}\right)^{2}=-\frac{2}{21}+\left(-\frac{17}{42}\right)^{2}$$

Square -\frac{17}{42} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{21}x+\frac{289}{1764}=-\frac{2}{21}+\frac{289}{1764}$$

Add -\frac{2}{21} to \frac{289}{1764} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{17}{21}x+\frac{289}{1764}=\frac{121}{1764}$$

Factor x^{2}-\frac{17}{21}x+\frac{289}{1764}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{42}\right)^{2}=\frac{121}{1764}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{42}\right)^{2}}=\sqrt{\frac{121}{1764}}$$

Simplify.

$$x-\frac{17}{42}=\frac{11}{42} x-\frac{17}{42}=-\frac{11}{42}$$

Add \frac{17}{42} to both sides of the equation.

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