Solve -(7x-1)^2+14=10

$$- ( 7 x - 1 ) ^ { 2 } + 14 = 10$$

Solve for x

$$x=\frac{3}{7}\approx 0.428571429$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-1\right)^{2}.

$$-\left(49x^{2}-14x+1\right)+14=10$$

To find the opposite of 49x^{2}-14x+1, find the opposite of each term.

$$-49x^{2}+14x-1+14=10$$

Add -1 and 14 to get 13.

$$-49x^{2}+14x+13=10$$

Subtract 10 from both sides.

$$-49x^{2}+14x+13-10=0$$

Subtract 10 from 13 to get 3.

$$-49x^{2}+14x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -49x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=14 ab=-49\times 3=-147$$

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. List all such integer pairs that give product -147.

$$-1,147 -3,49 -7,21$$

Calculate the sum for each pair.

$$-1+147=146 -3+49=46 -7+21=14$$

The solution is the pair that gives sum 14.

$$a=21 b=-7$$

Rewrite -49x^{2}+14x+3 as \left(-49x^{2}+21x\right)+\left(-7x+3\right).

$$\left(-49x^{2}+21x\right)+\left(-7x+3\right)$$

Factor out -7x in the first and -1 in the second group.

$$-7x\left(7x-3\right)-\left(7x-3\right)$$

Factor out common term 7x-3 by using distributive property.

$$\left(7x-3\right)\left(-7x-1\right)$$

To find equation solutions, solve 7x-3=0 and -7x-1=0.

$$x=\frac{3}{7} x=-\frac{1}{7}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-1\right)^{2}.

$$-\left(49x^{2}-14x+1\right)+14=10$$

To find the opposite of 49x^{2}-14x+1, find the opposite of each term.

$$-49x^{2}+14x-1+14=10$$

Add -1 and 14 to get 13.

$$-49x^{2}+14x+13=10$$

Subtract 10 from both sides.

$$-49x^{2}+14x+13-10=0$$

Subtract 10 from 13 to get 3.

$$-49x^{2}+14x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -49 for a, 14 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\left(-49\right)\times 3}}{2\left(-49\right)}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\left(-49\right)\times 3}}{2\left(-49\right)}$$

Multiply -4 times -49.

$$x=\frac{-14±\sqrt{196+196\times 3}}{2\left(-49\right)}$$

Multiply 196 times 3.

$$x=\frac{-14±\sqrt{196+588}}{2\left(-49\right)}$$

Add 196 to 588.

$$x=\frac{-14±\sqrt{784}}{2\left(-49\right)}$$

Take the square root of 784.

$$x=\frac{-14±28}{2\left(-49\right)}$$

Multiply 2 times -49.

$$x=\frac{-14±28}{-98}$$

Now solve the equation x=\frac{-14±28}{-98} when ± is plus. Add -14 to 28.

$$x=\frac{14}{-98}$$

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

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

Now solve the equation x=\frac{-14±28}{-98} when ± is minus. Subtract 28 from -14.

$$x=-\frac{42}{-98}$$

Reduce the fraction \frac{-42}{-98} to lowest terms by extracting and canceling out 14.

$$x=\frac{3}{7}$$

The equation is now solved.

$$x=-\frac{1}{7} x=\frac{3}{7}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-1\right)^{2}.

$$-\left(49x^{2}-14x+1\right)+14=10$$

To find the opposite of 49x^{2}-14x+1, find the opposite of each term.

$$-49x^{2}+14x-1+14=10$$

Add -1 and 14 to get 13.

$$-49x^{2}+14x+13=10$$

Subtract 13 from both sides.

$$-49x^{2}+14x=10-13$$

Subtract 13 from 10 to get -3.

$$-49x^{2}+14x=-3$$

Divide both sides by -49.

$$\frac{-49x^{2}+14x}{-49}=-\frac{3}{-49}$$

Dividing by -49 undoes the multiplication by -49.

$$x^{2}+\frac{14}{-49}x=-\frac{3}{-49}$$

Reduce the fraction \frac{14}{-49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{2}{7}x=-\frac{3}{-49}$$

Divide -3 by -49.

$$x^{2}-\frac{2}{7}x=\frac{3}{49}$$

Divide -\frac{2}{7}, the coefficient of the x term, by 2 to get -\frac{1}{7}. Then add the square of -\frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{7}x+\left(-\frac{1}{7}\right)^{2}=\frac{3}{49}+\left(-\frac{1}{7}\right)^{2}$$

Square -\frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{3+1}{49}$$

Add \frac{3}{49} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{7}x+\frac{1}{49}=\frac{4}{49}$$

Factor x^{2}-\frac{2}{7}x+\frac{1}{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}.

$$\left(x-\frac{1}{7}\right)^{2}=\frac{4}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{7}\right)^{2}}=\sqrt{\frac{4}{49}}$$

Simplify.

$$x-\frac{1}{7}=\frac{2}{7} x-\frac{1}{7}=-\frac{2}{7}$$

Add \frac{1}{7} to both sides of the equation.

$$x=\frac{3}{7} x=-\frac{1}{7}$$