Solve 7x^2+2=8x

$$7 { x }^{ 2 } +2=8x$$

Solve for x

$$x=\frac{\sqrt{2}+4}{7}\approx 0.77345908$$

$$x=\frac{4-\sqrt{2}}{7}\approx 0.369398063$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$7x^{2}+2-8x=0$$

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.

$$7x^{2}-8x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, -8 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7\times 2}}{2\times 7}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 7\times 2}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-8\right)±\sqrt{64-28\times 2}}{2\times 7}$$

Multiply -28 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-56}}{2\times 7}$$

Add 64 to -56.

$$x=\frac{-\left(-8\right)±\sqrt{8}}{2\times 7}$$

Take the square root of 8.

$$x=\frac{-\left(-8\right)±2\sqrt{2}}{2\times 7}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{2}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{8±2\sqrt{2}}{14}$$

Now solve the equation x=\frac{8±2\sqrt{2}}{14} when ± is plus. Add 8 to 2\sqrt{2}.

$$x=\frac{2\sqrt{2}+8}{14}$$

Divide 2\sqrt{2}+8 by 14.

$$x=\frac{\sqrt{2}+4}{7}$$

Now solve the equation x=\frac{8±2\sqrt{2}}{14} when ± is minus. Subtract 2\sqrt{2} from 8.

$$x=\frac{8-2\sqrt{2}}{14}$$

Divide 8-2\sqrt{2} by 14.

$$x=\frac{4-\sqrt{2}}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{2}+4}{7} x=\frac{4-\sqrt{2}}{7}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$7x^{2}+2-8x=0$$

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

$$7x^{2}-8x=-2$$

Divide both sides by 7.

$$\frac{7x^{2}-8x}{7}=-\frac{2}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{8}{7}x=-\frac{2}{7}$$

Divide -\frac{8}{7}, the coefficient of the x term, by 2 to get -\frac{4}{7}. Then add the square of -\frac{4}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{7}x+\left(-\frac{4}{7}\right)^{2}=-\frac{2}{7}+\left(-\frac{4}{7}\right)^{2}$$

Square -\frac{4}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{7}x+\frac{16}{49}=-\frac{2}{7}+\frac{16}{49}$$

Add -\frac{2}{7} to \frac{16}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{7}x+\frac{16}{49}=\frac{2}{49}$$

Factor x^{2}-\frac{8}{7}x+\frac{16}{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}.

$$\left(x-\frac{4}{7}\right)^{2}=\frac{2}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{7}\right)^{2}}=\sqrt{\frac{2}{49}}$$

Simplify.

$$x-\frac{4}{7}=\frac{\sqrt{2}}{7} x-\frac{4}{7}=-\frac{\sqrt{2}}{7}$$

Add \frac{4}{7} to both sides of the equation.

$$x=\frac{\sqrt{2}+4}{7} x=\frac{4-\sqrt{2}}{7}$$