Solve left(2x-3right)^2-left(x-5right)^2=-23

$${ \left(2x-3 \right) }^{ 2 } - { \left(x-5 \right) }^{ 2 } =-23$$

Solve for x (complex solution)

$$x=\frac{1+2\sqrt{5}i}{3}\approx 0.333333333+1.490711985i$$

$$x=\frac{-2\sqrt{5}i+1}{3}\approx 0.333333333-1.490711985i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-\left(x-5\right)^{2}=-23$$

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

$$4x^{2}-12x+9-\left(x^{2}-10x+25\right)=-23$$

To find the opposite of x^{2}-10x+25, find the opposite of each term.

$$4x^{2}-12x+9-x^{2}+10x-25=-23$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x+9+10x-25=-23$$

Combine -12x and 10x to get -2x.

$$3x^{2}-2x+9-25=-23$$

Subtract 25 from 9 to get -16.

$$3x^{2}-2x-16=-23$$

Add 23 to both sides.

$$3x^{2}-2x-16+23=0$$

Add -16 and 23 to get 7.

$$3x^{2}-2x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$x=\frac{-\left(-2\right)±\sqrt{4-84}}{2\times 3}$$

Add 4 to -84.

$$x=\frac{-\left(-2\right)±\sqrt{-80}}{2\times 3}$$

Take the square root of -80.

$$x=\frac{-\left(-2\right)±4\sqrt{5}i}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±4\sqrt{5}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±4\sqrt{5}i}{6}$$

Now solve the equation x=\frac{2±4\sqrt{5}i}{6} when ± is plus. Add 2 to 4i\sqrt{5}.

$$x=\frac{2+4\sqrt{5}i}{6}$$

Divide 2+4i\sqrt{5} by 6.

$$x=\frac{1+2\sqrt{5}i}{3}$$

Now solve the equation x=\frac{2±4\sqrt{5}i}{6} when ± is minus. Subtract 4i\sqrt{5} from 2.

$$x=\frac{-4\sqrt{5}i+2}{6}$$

Divide 2-4i\sqrt{5} by 6.

$$x=\frac{-2\sqrt{5}i+1}{3}$$

The equation is now solved.

$$x=\frac{1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i+1}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$4x^{2}-12x+9-\left(x-5\right)^{2}=-23$$

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

$$4x^{2}-12x+9-\left(x^{2}-10x+25\right)=-23$$

To find the opposite of x^{2}-10x+25, find the opposite of each term.

$$4x^{2}-12x+9-x^{2}+10x-25=-23$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-12x+9+10x-25=-23$$

Combine -12x and 10x to get -2x.

$$3x^{2}-2x+9-25=-23$$

Subtract 25 from 9 to get -16.

$$3x^{2}-2x-16=-23$$

Add 16 to both sides.

$$3x^{2}-2x=-23+16$$

Add -23 and 16 to get -7.

$$3x^{2}-2x=-7$$

Divide both sides by 3.

$$\frac{3x^{2}-2x}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{2}{3}x=-\frac{7}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{7}{3}+\frac{1}{9}$$

Add -\frac{7}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{20}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=-\frac{20}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{-\frac{20}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{2\sqrt{5}i}{3} x-\frac{1}{3}=-\frac{2\sqrt{5}i}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{1+2\sqrt{5}i}{3} x=\frac{-2\sqrt{5}i+1}{3}$$