Solve (2x+3^2)-(2x-1)(2x+1)=46

$$( 2 x + 3 ^ { 2 } ) - ( 2 x - 1 ) ( 2 x + 1 ) = 46$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{143}i+1}{4}\approx 0.25-2.989565186i$$

$$x=\frac{1+\sqrt{143}i}{4}\approx 0.25+2.989565186i$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$2x+9-\left(2x-1\right)\left(2x+1\right)=46$$

Consider \left(2x-1\right)\left(2x+1\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 1.

$$2x+9-\left(\left(2x\right)^{2}-1\right)=46$$

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

$$2x+9-\left(2^{2}x^{2}-1\right)=46$$

Calculate 2 to the power of 2 and get 4.

$$2x+9-\left(4x^{2}-1\right)=46$$

To find the opposite of 4x^{2}-1, find the opposite of each term.

$$2x+9-4x^{2}+1=46$$

Add 9 and 1 to get 10.

$$2x+10-4x^{2}=46$$

Subtract 46 from both sides.

$$2x+10-4x^{2}-46=0$$

Subtract 46 from 10 to get -36.

$$2x-36-4x^{2}=0$$

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 2 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-4\right)\left(-36\right)}}{2\left(-4\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-4\right)\left(-36\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16\left(-36\right)}}{2\left(-4\right)}$$

Multiply 16 times -36.

$$x=\frac{-2±\sqrt{4-576}}{2\left(-4\right)}$$

Add 4 to -576.

$$x=\frac{-2±\sqrt{-572}}{2\left(-4\right)}$$

Take the square root of -572.

$$x=\frac{-2±2\sqrt{143}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±2\sqrt{143}i}{-8}$$

Now solve the equation x=\frac{-2±2\sqrt{143}i}{-8} when ± is plus. Add -2 to 2i\sqrt{143}.

$$x=\frac{-2+2\sqrt{143}i}{-8}$$

Divide -2+2i\sqrt{143} by -8.

$$x=\frac{-\sqrt{143}i+1}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{143}i}{-8} when ± is minus. Subtract 2i\sqrt{143} from -2.

$$x=\frac{-2\sqrt{143}i-2}{-8}$$

Divide -2-2i\sqrt{143} by -8.

$$x=\frac{1+\sqrt{143}i}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{143}i+1}{4} x=\frac{1+\sqrt{143}i}{4}$$

Steps for Completing the Square

Calculate 3 to the power of 2 and get 9.

$$2x+9-\left(2x-1\right)\left(2x+1\right)=46$$

Consider \left(2x-1\right)\left(2x+1\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 1.

$$2x+9-\left(\left(2x\right)^{2}-1\right)=46$$

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

$$2x+9-\left(2^{2}x^{2}-1\right)=46$$

Calculate 2 to the power of 2 and get 4.

$$2x+9-\left(4x^{2}-1\right)=46$$

To find the opposite of 4x^{2}-1, find the opposite of each term.

$$2x+9-4x^{2}+1=46$$

Add 9 and 1 to get 10.

$$2x+10-4x^{2}=46$$

Subtract 10 from both sides.

$$2x-4x^{2}=46-10$$

Subtract 10 from 46 to get 36.

$$2x-4x^{2}=36$$

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.

$$-4x^{2}+2x=36$$

Divide both sides by -4.

$$\frac{-4x^{2}+2x}{-4}=\frac{36}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2}{-4}x=\frac{36}{-4}$$

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

$$x^{2}-\frac{1}{2}x=\frac{36}{-4}$$

Divide 36 by -4.

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

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-9+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-9+\frac{1}{16}$$

Add -9 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{143}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{143}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{143}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{143}i}{4} x-\frac{1}{4}=-\frac{\sqrt{143}i}{4}$$

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

$$x=\frac{1+\sqrt{143}i}{4} x=\frac{-\sqrt{143}i+1}{4}$$