Solve (2x-7)^2-7(7-4x)=0

$$( 2 x - 7 ) ^ { 2 } - 7 ( 7 - 4 x ) = 0$$

Solve for x

$$x=0$$

Steps by Finding Square Root

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

$$4x^{2}-28x+49-7\left(7-4x\right)=0$$

Use the distributive property to multiply -7 by 7-4x.

$$4x^{2}-28x+49-49+28x=0$$

Subtract 49 from 49 to get 0.

$$4x^{2}-28x+28x=0$$

Combine -28x and 28x to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

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

$$4x^{2}-28x+49-7\left(7-4x\right)=0$$

Use the distributive property to multiply -7 by 7-4x.

$$4x^{2}-28x+49-49+28x=0$$

Subtract 49 from 49 to get 0.

$$4x^{2}-28x+28x=0$$

Combine -28x and 28x to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

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

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$