Solve (7x-2)^2-1=80

$$( 7 x - 2 ) ^ { 2 } - 1 = 80$$

Solve for x

$$x = \frac{11}{7} = 1\frac{4}{7} \approx 1.571428571$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-2\right)^{2}.

$$49x^{2}-28x+4-1=80$$

Subtract 1 from 4 to get 3.

$$49x^{2}-28x+3=80$$

Subtract 80 from both sides.

$$49x^{2}-28x+3-80=0$$

Subtract 80 from 3 to get -77.

$$49x^{2}-28x-77=0$$

Divide both sides by 7.

$$7x^{2}-4x-11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-11. To find a and b, set up a system to be solved.

$$a+b=-4 ab=7\left(-11\right)=-77$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -77.

$$1,-77 7,-11$$

Calculate the sum for each pair.

$$1-77=-76 7-11=-4$$

The solution is the pair that gives sum -4.

$$a=-11 b=7$$

Rewrite 7x^{2}-4x-11 as \left(7x^{2}-11x\right)+\left(7x-11\right).

$$\left(7x^{2}-11x\right)+\left(7x-11\right)$$

Factor out x in 7x^{2}-11x.

$$x\left(7x-11\right)+7x-11$$

Factor out common term 7x-11 by using distributive property.

$$\left(7x-11\right)\left(x+1\right)$$

To find equation solutions, solve 7x-11=0 and x+1=0.

$$x=\frac{11}{7} x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-2\right)^{2}.

$$49x^{2}-28x+4-1=80$$

Subtract 1 from 4 to get 3.

$$49x^{2}-28x+3=80$$

Subtract 80 from both sides.

$$49x^{2}-28x+3-80=0$$

Subtract 80 from 3 to get -77.

$$49x^{2}-28x-77=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -28 for b, and -77 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 49\left(-77\right)}}{2\times 49}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 49\left(-77\right)}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-28\right)±\sqrt{784-196\left(-77\right)}}{2\times 49}$$

Multiply -196 times -77.

$$x=\frac{-\left(-28\right)±\sqrt{784+15092}}{2\times 49}$$

Add 784 to 15092.

$$x=\frac{-\left(-28\right)±\sqrt{15876}}{2\times 49}$$

Take the square root of 15876.

$$x=\frac{-\left(-28\right)±126}{2\times 49}$$

The opposite of -28 is 28.

$$x=\frac{28±126}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{28±126}{98}$$

Now solve the equation x=\frac{28±126}{98} when ± is plus. Add 28 to 126.

$$x=\frac{154}{98}$$

Reduce the fraction \frac{154}{98} to lowest terms by extracting and canceling out 14.

$$x=\frac{11}{7}$$

Now solve the equation x=\frac{28±126}{98} when ± is minus. Subtract 126 from 28.

$$x=-\frac{98}{98}$$

Divide -98 by 98.

$$x=-1$$

The equation is now solved.

$$x=\frac{11}{7} x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-2\right)^{2}.

$$49x^{2}-28x+4-1=80$$

Subtract 1 from 4 to get 3.

$$49x^{2}-28x+3=80$$

Subtract 3 from both sides.

$$49x^{2}-28x=80-3$$

Subtract 3 from 80 to get 77.

$$49x^{2}-28x=77$$

Divide both sides by 49.

$$\frac{49x^{2}-28x}{49}=\frac{77}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\left(-\frac{28}{49}\right)x=\frac{77}{49}$$

Reduce the fraction \frac{-28}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{4}{7}x=\frac{77}{49}$$

Reduce the fraction \frac{77}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{4}{7}x=\frac{11}{7}$$

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

$$x^{2}-\frac{4}{7}x+\left(-\frac{2}{7}\right)^{2}=\frac{11}{7}+\left(-\frac{2}{7}\right)^{2}$$

Square -\frac{2}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=\frac{11}{7}+\frac{4}{49}$$

Add \frac{11}{7} to \frac{4}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=\frac{81}{49}$$

Factor x^{2}-\frac{4}{7}x+\frac{4}{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}.

$$\left(x-\frac{2}{7}\right)^{2}=\frac{81}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{7}\right)^{2}}=\sqrt{\frac{81}{49}}$$

Simplify.

$$x-\frac{2}{7}=\frac{9}{7} x-\frac{2}{7}=-\frac{9}{7}$$

Add \frac{2}{7} to both sides of the equation.

$$x=\frac{11}{7} x=-1$$