Solve 1=7x^2+6x

$$1 = 7 x ^ { 2 } + 6 x$$

Solve for x

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

$$x=-1$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$7x^{2}+6x=1$$

Subtract 1 from both sides.

$$7x^{2}+6x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=6 ab=7\left(-1\right)=-7$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=7$$

Rewrite 7x^{2}+6x-1 as \left(7x^{2}-x\right)+\left(7x-1\right).

$$\left(7x^{2}-x\right)+\left(7x-1\right)$$

Factor out x in 7x^{2}-x.

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

Factor out common term 7x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$7x^{2}+6x=1$$

Subtract 1 from both sides.

$$7x^{2}+6x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 6 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 7\left(-1\right)}}{2\times 7}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 7\left(-1\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-6±\sqrt{36-28\left(-1\right)}}{2\times 7}$$

Multiply -28 times -1.

$$x=\frac{-6±\sqrt{36+28}}{2\times 7}$$

Add 36 to 28.

$$x=\frac{-6±\sqrt{64}}{2\times 7}$$

Take the square root of 64.

$$x=\frac{-6±8}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-6±8}{14}$$

Now solve the equation x=\frac{-6±8}{14} when ± is plus. Add -6 to 8.

$$x=\frac{2}{14}$$

Reduce the fraction \frac{2}{14} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-6±8}{14} when ± is minus. Subtract 8 from -6.

$$x=-\frac{14}{14}$$

Divide -14 by 14.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$7x^{2}+6x=1$$

Divide both sides by 7.

$$\frac{7x^{2}+6x}{7}=\frac{1}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{6}{7}x=\frac{1}{7}$$

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.

$$x^{2}+\frac{6}{7}x+\left(\frac{3}{7}\right)^{2}=\frac{1}{7}+\left(\frac{3}{7}\right)^{2}$$

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

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=\frac{1}{7}+\frac{9}{49}$$

Add \frac{1}{7} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=\frac{16}{49}$$

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}.

$$\left(x+\frac{3}{7}\right)^{2}=\frac{16}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{7}\right)^{2}}=\sqrt{\frac{16}{49}}$$

Simplify.

$$x+\frac{3}{7}=\frac{4}{7} x+\frac{3}{7}=-\frac{4}{7}$$

Subtract \frac{3}{7} from both sides of the equation.

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