Solve 2x-5^2-left(2x-3right)left(2x+3right)=0

$$2x- { 5 }^{ 2 } - \left( 2x-3 \right) \left( 2x+3 \right) = 0$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{7}i+1}{4}\approx 0.25-1.984313483i$$

$$x=\frac{1+3\sqrt{7}i}{4}\approx 0.25+1.984313483i$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$2x-25-\left(2x-3\right)\left(2x+3\right)=0$$

Consider \left(2x-3\right)\left(2x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4x^{2}-9, find the opposite of each term.

$$2x-25-4x^{2}+9=0$$

Add -25 and 9 to get -16.

$$2x-16-4x^{2}=0$$

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.

$$-4x^{2}+2x-16=0$$

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

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

Square 2.

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

Multiply -4 times -4.

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

Multiply 16 times -16.

$$x=\frac{-2±\sqrt{4-256}}{2\left(-4\right)}$$

Add 4 to -256.

$$x=\frac{-2±\sqrt{-252}}{2\left(-4\right)}$$

Take the square root of -252.

$$x=\frac{-2±6\sqrt{7}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±6\sqrt{7}i}{-8}$$

Now solve the equation x=\frac{-2±6\sqrt{7}i}{-8} when ± is plus. Add -2 to 6i\sqrt{7}.

$$x=\frac{-2+6\sqrt{7}i}{-8}$$

Divide -2+6i\sqrt{7} by -8.

$$x=\frac{-3\sqrt{7}i+1}{4}$$

Now solve the equation x=\frac{-2±6\sqrt{7}i}{-8} when ± is minus. Subtract 6i\sqrt{7} from -2.

$$x=\frac{-6\sqrt{7}i-2}{-8}$$

Divide -2-6i\sqrt{7} by -8.

$$x=\frac{1+3\sqrt{7}i}{4}$$

The equation is now solved.

$$x=\frac{-3\sqrt{7}i+1}{4} x=\frac{1+3\sqrt{7}i}{4}$$

Steps for Completing the Square

Calculate 5 to the power of 2 and get 25.

$$2x-25-\left(2x-3\right)\left(2x+3\right)=0$$

Consider \left(2x-3\right)\left(2x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

Expand \left(2x\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4x^{2}-9, find the opposite of each term.

$$2x-25-4x^{2}+9=0$$

Add -25 and 9 to get -16.

$$2x-16-4x^{2}=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$2x-4x^{2}=16$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}+2x=16$$

Divide both sides by -4.

$$\frac{-4x^{2}+2x}{-4}=\frac{16}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2}{-4}x=\frac{16}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=\frac{16}{-4}$$

Divide 16 by -4.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-4+\frac{1}{16}$$

Add -4 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{63}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{63}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{63}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{3\sqrt{7}i}{4} x-\frac{1}{4}=-\frac{3\sqrt{7}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+3\sqrt{7}i}{4} x=\frac{-3\sqrt{7}i+1}{4}$$