Solve (7/2x-1)^2=-7x+26

$$( \frac { 7 } { 2 } x - 1 ) ^ { 2 } = - 7 x + 26$$

Solve for x

$$x = -\frac{10}{7} = -1\frac{3}{7} \approx -1.428571429$$

$$x = \frac{10}{7} = 1\frac{3}{7} \approx 1.428571429$$

Steps Using Difference of Squares

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

$$\frac{49}{4}x^{2}-7x+1=-7x+26$$

Add 7x to both sides.

$$\frac{49}{4}x^{2}-7x+1+7x=26$$

Combine -7x and 7x to get 0.

$$\frac{49}{4}x^{2}+1=26$$

Subtract 26 from both sides.

$$\frac{49}{4}x^{2}+1-26=0$$

Subtract 26 from 1 to get -25.

$$\frac{49}{4}x^{2}-25=0$$

Multiply both sides by 4.

$$49x^{2}-100=0$$

Consider 49x^{2}-100. Rewrite 49x^{2}-100 as \left(7x\right)^{2}-10^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(7x-10\right)\left(7x+10\right)=0$$

To find equation solutions, solve 7x-10=0 and 7x+10=0.

$$x=\frac{10}{7} x=-\frac{10}{7}$$

Steps by Finding Square Root

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

$$\frac{49}{4}x^{2}-7x+1=-7x+26$$

Add 7x to both sides.

$$\frac{49}{4}x^{2}-7x+1+7x=26$$

Combine -7x and 7x to get 0.

$$\frac{49}{4}x^{2}+1=26$$

Subtract 1 from both sides.

$$\frac{49}{4}x^{2}=26-1$$

Subtract 1 from 26 to get 25.

$$\frac{49}{4}x^{2}=25$$

Multiply both sides by \frac{4}{49}, the reciprocal of \frac{49}{4}.

$$x^{2}=25\times \frac{4}{49}$$

Multiply 25 and \frac{4}{49} to get \frac{100}{49}.

$$x^{2}=\frac{100}{49}$$

Take the square root of both sides of the equation.

$$x=\frac{10}{7} x=-\frac{10}{7}$$

Steps Using the Quadratic Formula

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

$$\frac{49}{4}x^{2}-7x+1=-7x+26$$

Add 7x to both sides.

$$\frac{49}{4}x^{2}-7x+1+7x=26$$

Combine -7x and 7x to get 0.

$$\frac{49}{4}x^{2}+1=26$$

Subtract 26 from both sides.

$$\frac{49}{4}x^{2}+1-26=0$$

Subtract 26 from 1 to get -25.

$$\frac{49}{4}x^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{49}{4} for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{49}{4}\left(-25\right)}}{2\times \frac{49}{4}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{49}{4}\left(-25\right)}}{2\times \frac{49}{4}}$$

Multiply -4 times \frac{49}{4}.

$$x=\frac{0±\sqrt{-49\left(-25\right)}}{2\times \frac{49}{4}}$$

Multiply -49 times -25.

$$x=\frac{0±\sqrt{1225}}{2\times \frac{49}{4}}$$

Take the square root of 1225.

$$x=\frac{0±35}{2\times \frac{49}{4}}$$

Multiply 2 times \frac{49}{4}.

$$x=\frac{0±35}{\frac{49}{2}}$$

Now solve the equation x=\frac{0±35}{\frac{49}{2}} when ± is plus. Divide 35 by \frac{49}{2} by multiplying 35 by the reciprocal of \frac{49}{2}.

$$x=\frac{10}{7}$$

Now solve the equation x=\frac{0±35}{\frac{49}{2}} when ± is minus. Divide -35 by \frac{49}{2} by multiplying -35 by the reciprocal of \frac{49}{2}.

$$x=-\frac{10}{7}$$

The equation is now solved.

$$x=\frac{10}{7} x=-\frac{10}{7}$$