Solve 6x^2-8x+1=x+4

$$6 x ^ { 2 } - 8 x + 1 = x + 4$$

Solve for x

$$x = \frac{\sqrt{17} + 3}{4} \approx 1.780776406$$

$$x=\frac{3-\sqrt{17}}{4}\approx -0.280776406$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6x^{2}-8x+1-x=4$$

Combine -8x and -x to get -9x.

$$6x^{2}-9x+1=4$$

Subtract 4 from both sides.

$$6x^{2}-9x+1-4=0$$

Subtract 4 from 1 to get -3.

$$6x^{2}-9x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -9 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 6\left(-3\right)}}{2\times 6}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 6\left(-3\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-9\right)±\sqrt{81-24\left(-3\right)}}{2\times 6}$$

Multiply -24 times -3.

$$x=\frac{-\left(-9\right)±\sqrt{81+72}}{2\times 6}$$

Add 81 to 72.

$$x=\frac{-\left(-9\right)±\sqrt{153}}{2\times 6}$$

Take the square root of 153.

$$x=\frac{-\left(-9\right)±3\sqrt{17}}{2\times 6}$$

The opposite of -9 is 9.

$$x=\frac{9±3\sqrt{17}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{9±3\sqrt{17}}{12}$$

Now solve the equation x=\frac{9±3\sqrt{17}}{12} when ± is plus. Add 9 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}+9}{12}$$

Divide 9+3\sqrt{17} by 12.

$$x=\frac{\sqrt{17}+3}{4}$$

Now solve the equation x=\frac{9±3\sqrt{17}}{12} when ± is minus. Subtract 3\sqrt{17} from 9.

$$x=\frac{9-3\sqrt{17}}{12}$$

Divide 9-3\sqrt{17} by 12.

$$x=\frac{3-\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$

Steps for Completing the Square

Subtract x from both sides.

$$6x^{2}-8x+1-x=4$$

Combine -8x and -x to get -9x.

$$6x^{2}-9x+1=4$$

Subtract 1 from both sides.

$$6x^{2}-9x=4-1$$

Subtract 1 from 4 to get 3.

$$6x^{2}-9x=3$$

Divide both sides by 6.

$$\frac{6x^{2}-9x}{6}=\frac{3}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{9}{6}\right)x=\frac{3}{6}$$

Reduce the fraction \frac{-9}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{3}{2}x=\frac{3}{6}$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{3}{2}x=\frac{1}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Add \frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{17}}{4} x-\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$