Solve (1-3x)(2x-3)=4x^2+1

$$( 1 - 3 x ) ( 2 x - 3 ) = 4 x ^ { 2 } + 1$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{39}i+11}{20}\approx 0.55-0.3122499i$$

$$x=\frac{11+\sqrt{39}i}{20}\approx 0.55+0.3122499i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 1-3x by 2x-3 and combine like terms.

$$11x-3-6x^{2}=4x^{2}+1$$

Subtract 4x^{2} from both sides.

$$11x-3-6x^{2}-4x^{2}=1$$

Combine -6x^{2} and -4x^{2} to get -10x^{2}.

$$11x-3-10x^{2}=1$$

Subtract 1 from both sides.

$$11x-3-10x^{2}-1=0$$

Subtract 1 from -3 to get -4.

$$11x-4-10x^{2}=0$$

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.

$$-10x^{2}+11x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 11 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-10\right)\left(-4\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-11±\sqrt{121+40\left(-4\right)}}{2\left(-10\right)}$$

Multiply 40 times -4.

$$x=\frac{-11±\sqrt{121-160}}{2\left(-10\right)}$$

Add 121 to -160.

$$x=\frac{-11±\sqrt{-39}}{2\left(-10\right)}$$

Take the square root of -39.

$$x=\frac{-11±\sqrt{39}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{-11±\sqrt{39}i}{-20}$$

Now solve the equation x=\frac{-11±\sqrt{39}i}{-20} when ± is plus. Add -11 to i\sqrt{39}.

$$x=\frac{-11+\sqrt{39}i}{-20}$$

Divide -11+i\sqrt{39} by -20.

$$x=\frac{-\sqrt{39}i+11}{20}$$

Now solve the equation x=\frac{-11±\sqrt{39}i}{-20} when ± is minus. Subtract i\sqrt{39} from -11.

$$x=\frac{-\sqrt{39}i-11}{-20}$$

Divide -11-i\sqrt{39} by -20.

$$x=\frac{11+\sqrt{39}i}{20}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i+11}{20} x=\frac{11+\sqrt{39}i}{20}$$

Steps for Completing the Square

Use the distributive property to multiply 1-3x by 2x-3 and combine like terms.

$$11x-3-6x^{2}=4x^{2}+1$$

Subtract 4x^{2} from both sides.

$$11x-3-6x^{2}-4x^{2}=1$$

Combine -6x^{2} and -4x^{2} to get -10x^{2}.

$$11x-3-10x^{2}=1$$

Add 3 to both sides.

$$11x-10x^{2}=1+3$$

Add 1 and 3 to get 4.

$$11x-10x^{2}=4$$

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.

$$-10x^{2}+11x=4$$

Divide both sides by -10.

$$\frac{-10x^{2}+11x}{-10}=\frac{4}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{11}{-10}x=\frac{4}{-10}$$

Divide 11 by -10.

$$x^{2}-\frac{11}{10}x=\frac{4}{-10}$$

Reduce the fraction \frac{4}{-10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{10}x=-\frac{2}{5}$$

Divide -\frac{11}{10}, the coefficient of the x term, by 2 to get -\frac{11}{20}. Then add the square of -\frac{11}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{10}x+\left(-\frac{11}{20}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{20}\right)^{2}$$

Square -\frac{11}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{10}x+\frac{121}{400}=-\frac{2}{5}+\frac{121}{400}$$

Add -\frac{2}{5} to \frac{121}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{10}x+\frac{121}{400}=-\frac{39}{400}$$

Factor x^{2}-\frac{11}{10}x+\frac{121}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{20}\right)^{2}=-\frac{39}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{20}\right)^{2}}=\sqrt{-\frac{39}{400}}$$

Simplify.

$$x-\frac{11}{20}=\frac{\sqrt{39}i}{20} x-\frac{11}{20}=-\frac{\sqrt{39}i}{20}$$

Add \frac{11}{20} to both sides of the equation.

$$x=\frac{11+\sqrt{39}i}{20} x=\frac{-\sqrt{39}i+11}{20}$$