Solve 4x^2-25-9(2x-5)^2=0

$$4 x ^ { 2 } - 25 - 9 ( 2 x - 5 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{25}{8} = 3\frac{1}{8} = 3.125$$

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

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}.

$$4x^{2}-25-9\left(4x^{2}-20x+25\right)=0$$

Use the distributive property to multiply -9 by 4x^{2}-20x+25.

$$4x^{2}-25-36x^{2}+180x-225=0$$

Combine 4x^{2} and -36x^{2} to get -32x^{2}.

$$-32x^{2}-25+180x-225=0$$

Subtract 225 from -25 to get -250.

$$-32x^{2}-250+180x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-32x^{2}+180x-250=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -32 for a, 180 for b, and -250 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-180±\sqrt{180^{2}-4\left(-32\right)\left(-250\right)}}{2\left(-32\right)}$$

Square 180.

$$x=\frac{-180±\sqrt{32400-4\left(-32\right)\left(-250\right)}}{2\left(-32\right)}$$

Multiply -4 times -32.

$$x=\frac{-180±\sqrt{32400+128\left(-250\right)}}{2\left(-32\right)}$$

Multiply 128 times -250.

$$x=\frac{-180±\sqrt{32400-32000}}{2\left(-32\right)}$$

Add 32400 to -32000.

$$x=\frac{-180±\sqrt{400}}{2\left(-32\right)}$$

Take the square root of 400.

$$x=\frac{-180±20}{2\left(-32\right)}$$

Multiply 2 times -32.

$$x=\frac{-180±20}{-64}$$

Now solve the equation x=\frac{-180±20}{-64} when ± is plus. Add -180 to 20.

$$x=-\frac{160}{-64}$$

Reduce the fraction \frac{-160}{-64} to lowest terms by extracting and canceling out 32.

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

Now solve the equation x=\frac{-180±20}{-64} when ± is minus. Subtract 20 from -180.

$$x=-\frac{200}{-64}$$

Reduce the fraction \frac{-200}{-64} to lowest terms by extracting and canceling out 8.

$$x=\frac{25}{8}$$

The equation is now solved.

$$x=\frac{5}{2} x=\frac{25}{8}$$

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}.

$$4x^{2}-25-9\left(4x^{2}-20x+25\right)=0$$

Use the distributive property to multiply -9 by 4x^{2}-20x+25.

$$4x^{2}-25-36x^{2}+180x-225=0$$

Combine 4x^{2} and -36x^{2} to get -32x^{2}.

$$-32x^{2}-25+180x-225=0$$

Subtract 225 from -25 to get -250.

$$-32x^{2}-250+180x=0$$

Add 250 to both sides. Anything plus zero gives itself.

$$-32x^{2}+180x=250$$

Divide both sides by -32.

$$\frac{-32x^{2}+180x}{-32}=\frac{250}{-32}$$

Dividing by -32 undoes the multiplication by -32.

$$x^{2}+\frac{180}{-32}x=\frac{250}{-32}$$

Reduce the fraction \frac{180}{-32} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{45}{8}x=\frac{250}{-32}$$

Reduce the fraction \frac{250}{-32} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{45}{8}x=-\frac{125}{16}$$

Divide -\frac{45}{8}, the coefficient of the x term, by 2 to get -\frac{45}{16}. Then add the square of -\frac{45}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{45}{8}x+\left(-\frac{45}{16}\right)^{2}=-\frac{125}{16}+\left(-\frac{45}{16}\right)^{2}$$

Square -\frac{45}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{45}{8}x+\frac{2025}{256}=-\frac{125}{16}+\frac{2025}{256}$$

Add -\frac{125}{16} to \frac{2025}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{45}{8}x+\frac{2025}{256}=\frac{25}{256}$$

Factor x^{2}-\frac{45}{8}x+\frac{2025}{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}.

$$\left(x-\frac{45}{16}\right)^{2}=\frac{25}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{45}{16}\right)^{2}}=\sqrt{\frac{25}{256}}$$

Simplify.

$$x-\frac{45}{16}=\frac{5}{16} x-\frac{45}{16}=-\frac{5}{16}$$

Add \frac{45}{16} to both sides of the equation.

$$x=\frac{25}{8} x=\frac{5}{2}$$