Solve x^2-1-(x-1)(3x+5)=20

$$x ^ { 2 } - 1 - ( x - 1 ) ( 3 x + 5 ) = 20$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{31}i}{2}\approx -0.5+2.783882181i$$

$$x=\frac{-\sqrt{31}i-1}{2}\approx -0.5-2.783882181i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-1 by 3x+5 and combine like terms.

$$x^{2}-1-\left(3x^{2}+2x-5\right)=20$$

To find the opposite of 3x^{2}+2x-5, find the opposite of each term.

$$x^{2}-1-3x^{2}-2x+5=20$$

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

$$-2x^{2}-1-2x+5=20$$

Add -1 and 5 to get 4.

$$-2x^{2}+4-2x=20$$

Subtract 20 from both sides.

$$-2x^{2}+4-2x-20=0$$

Subtract 20 from 4 to get -16.

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

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.

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

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

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

Square -2.

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

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\left(-16\right)}}{2\left(-2\right)}$$

Multiply 8 times -16.

$$x=\frac{-\left(-2\right)±\sqrt{4-128}}{2\left(-2\right)}$$

Add 4 to -128.

$$x=\frac{-\left(-2\right)±\sqrt{-124}}{2\left(-2\right)}$$

Take the square root of -124.

$$x=\frac{-\left(-2\right)±2\sqrt{31}i}{2\left(-2\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{31}i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{31}i}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{31}i}{-4} when ± is plus. Add 2 to 2i\sqrt{31}.

$$x=\frac{2+2\sqrt{31}i}{-4}$$

Divide 2+2i\sqrt{31} by -4.

$$x=\frac{-\sqrt{31}i-1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{31}i}{-4} when ± is minus. Subtract 2i\sqrt{31} from 2.

$$x=\frac{-2\sqrt{31}i+2}{-4}$$

Divide 2-2i\sqrt{31} by -4.

$$x=\frac{-1+\sqrt{31}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{31}i-1}{2} x=\frac{-1+\sqrt{31}i}{2}$$

Steps for Completing the Square

Use the distributive property to multiply x-1 by 3x+5 and combine like terms.

$$x^{2}-1-\left(3x^{2}+2x-5\right)=20$$

To find the opposite of 3x^{2}+2x-5, find the opposite of each term.

$$x^{2}-1-3x^{2}-2x+5=20$$

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

$$-2x^{2}-1-2x+5=20$$

Add -1 and 5 to get 4.

$$-2x^{2}+4-2x=20$$

Subtract 4 from both sides.

$$-2x^{2}-2x=20-4$$

Subtract 4 from 20 to get 16.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -2 by -2.

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

Divide 16 by -2.

$$x^{2}+x=-8$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=-8+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=-8+\frac{1}{4}$$

Add -8 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{31}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{31}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{31}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{31}i}{2} x+\frac{1}{2}=-\frac{\sqrt{31}i}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{-1+\sqrt{31}i}{2} x=\frac{-\sqrt{31}i-1}{2}$$