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

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

Solve for x

$$x=\frac{\sqrt{39}}{21}+\frac{3}{7}\approx 0.725952286$$

$$x=-\frac{\sqrt{39}}{21}+\frac{3}{7}\approx 0.131190572$$

Steps Using the Quadratic Formula

Multiply 7 and 3 to get 21.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -18 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 21\times 2}}{2\times 21}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 21\times 2}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-\left(-18\right)±\sqrt{324-84\times 2}}{2\times 21}$$

Multiply -84 times 2.

$$x=\frac{-\left(-18\right)±\sqrt{324-168}}{2\times 21}$$

Add 324 to -168.

$$x=\frac{-\left(-18\right)±\sqrt{156}}{2\times 21}$$

Take the square root of 156.

$$x=\frac{-\left(-18\right)±2\sqrt{39}}{2\times 21}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{39}}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{18±2\sqrt{39}}{42}$$

Now solve the equation x=\frac{18±2\sqrt{39}}{42} when ± is plus. Add 18 to 2\sqrt{39}.

$$x=\frac{2\sqrt{39}+18}{42}$$

Divide 18+2\sqrt{39} by 42.

$$x=\frac{\sqrt{39}}{21}+\frac{3}{7}$$

Now solve the equation x=\frac{18±2\sqrt{39}}{42} when ± is minus. Subtract 2\sqrt{39} from 18.

$$x=\frac{18-2\sqrt{39}}{42}$$

Divide 18-2\sqrt{39} by 42.

$$x=-\frac{\sqrt{39}}{21}+\frac{3}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{39}}{21}+\frac{3}{7} x=-\frac{\sqrt{39}}{21}+\frac{3}{7}$$

Steps for Completing the Square

Multiply 7 and 3 to get 21.

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

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

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

Divide both sides by 21.

$$\frac{21x^{2}-18x}{21}=-\frac{2}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$x^{2}+\left(-\frac{18}{21}\right)x=-\frac{2}{21}$$

Reduce the fraction \frac{-18}{21} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{6}{7}x=-\frac{2}{21}$$

Divide -\frac{6}{7}, the coefficient of the x term, by 2 to get -\frac{3}{7}. Then add the square of -\frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{6}{7}x+\left(-\frac{3}{7}\right)^{2}=-\frac{2}{21}+\left(-\frac{3}{7}\right)^{2}$$

Square -\frac{3}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{6}{7}x+\frac{9}{49}=-\frac{2}{21}+\frac{9}{49}$$

Add -\frac{2}{21} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{7}x+\frac{9}{49}=\frac{13}{147}$$

Factor x^{2}-\frac{6}{7}x+\frac{9}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{7}\right)^{2}=\frac{13}{147}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{7}\right)^{2}}=\sqrt{\frac{13}{147}}$$

Simplify.

$$x-\frac{3}{7}=\frac{\sqrt{39}}{21} x-\frac{3}{7}=-\frac{\sqrt{39}}{21}$$

Add \frac{3}{7} to both sides of the equation.

$$x=\frac{\sqrt{39}}{21}+\frac{3}{7} x=-\frac{\sqrt{39}}{21}+\frac{3}{7}$$