Solve 4(2x-5)^2=5-2x

$$4 ( 2 x - 5 ) ^ { 2 } = 5 - 2 x$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = \frac{19}{8} = 2\frac{3}{8} = 2.375$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4\left(4x^{2}-20x+25\right)=5-2x$$

Use the distributive property to multiply 4 by 4x^{2}-20x+25.

$$16x^{2}-80x+100=5-2x$$

Subtract 5 from both sides.

$$16x^{2}-80x+100-5=-2x$$

Subtract 5 from 100 to get 95.

$$16x^{2}-80x+95=-2x$$

Add 2x to both sides.

$$16x^{2}-80x+95+2x=0$$

Combine -80x and 2x to get -78x.

$$16x^{2}-78x+95=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -78 for b, and 95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-78\right)±\sqrt{\left(-78\right)^{2}-4\times 16\times 95}}{2\times 16}$$

Square -78.

$$x=\frac{-\left(-78\right)±\sqrt{6084-4\times 16\times 95}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-78\right)±\sqrt{6084-64\times 95}}{2\times 16}$$

Multiply -64 times 95.

$$x=\frac{-\left(-78\right)±\sqrt{6084-6080}}{2\times 16}$$

Add 6084 to -6080.

$$x=\frac{-\left(-78\right)±\sqrt{4}}{2\times 16}$$

Take the square root of 4.

$$x=\frac{-\left(-78\right)±2}{2\times 16}$$

The opposite of -78 is 78.

$$x=\frac{78±2}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{78±2}{32}$$

Now solve the equation x=\frac{78±2}{32} when ± is plus. Add 78 to 2.

$$x=\frac{80}{32}$$

Reduce the fraction \frac{80}{32} to lowest terms by extracting and canceling out 16.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{78±2}{32} when ± is minus. Subtract 2 from 78.

$$x=\frac{76}{32}$$

Reduce the fraction \frac{76}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{19}{8}$$

The equation is now solved.

$$x=\frac{5}{2} x=\frac{19}{8}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4\left(4x^{2}-20x+25\right)=5-2x$$

Use the distributive property to multiply 4 by 4x^{2}-20x+25.

$$16x^{2}-80x+100=5-2x$$

Add 2x to both sides.

$$16x^{2}-80x+100+2x=5$$

Combine -80x and 2x to get -78x.

$$16x^{2}-78x+100=5$$

Subtract 100 from both sides.

$$16x^{2}-78x=5-100$$

Subtract 100 from 5 to get -95.

$$16x^{2}-78x=-95$$

Divide both sides by 16.

$$\frac{16x^{2}-78x}{16}=-\frac{95}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{78}{16}\right)x=-\frac{95}{16}$$

Reduce the fraction \frac{-78}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{39}{8}x=-\frac{95}{16}$$

Divide -\frac{39}{8}, the coefficient of the x term, by 2 to get -\frac{39}{16}. Then add the square of -\frac{39}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{39}{8}x+\left(-\frac{39}{16}\right)^{2}=-\frac{95}{16}+\left(-\frac{39}{16}\right)^{2}$$

Square -\frac{39}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{39}{8}x+\frac{1521}{256}=-\frac{95}{16}+\frac{1521}{256}$$

Add -\frac{95}{16} to \frac{1521}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{39}{8}x+\frac{1521}{256}=\frac{1}{256}$$

Factor x^{2}-\frac{39}{8}x+\frac{1521}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{39}{16}\right)^{2}=\frac{1}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{39}{16}\right)^{2}}=\sqrt{\frac{1}{256}}$$

Simplify.

$$x-\frac{39}{16}=\frac{1}{16} x-\frac{39}{16}=-\frac{1}{16}$$

Add \frac{39}{16} to both sides of the equation.

$$x=\frac{5}{2} x=\frac{19}{8}$$