Solve 3(x-1)-(x-3)div5(x-2)=18

$$3(x-1)-(x-3) \div 5(x-2)=18$$

Solve for x (complex solution)

$$x=10+\sqrt{11}i\approx 10+3.31662479i$$

$$x=-\sqrt{11}i+10\approx 10-3.31662479i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$15\left(x-1\right)-\left(x-3\right)\left(x-2\right)=90$$

Use the distributive property to multiply 15 by x-1.

$$15x-15-\left(x-3\right)\left(x-2\right)=90$$

Apply the distributive property by multiplying each term of x-3 by each term of x-2.

$$15x-15-\left(x^{2}-2x-3x+6\right)=90$$

Combine -2x and -3x to get -5x.

$$15x-15-\left(x^{2}-5x+6\right)=90$$

To find the opposite of x^{2}-5x+6, find the opposite of each term.

$$15x-15-x^{2}-\left(-5x\right)-6=90$$

The opposite of -5x is 5x.

$$15x-15-x^{2}+5x-6=90$$

Combine 15x and 5x to get 20x.

$$20x-15-x^{2}-6=90$$

Subtract 6 from -15 to get -21.

$$20x-21-x^{2}=90$$

Subtract 90 from both sides.

$$20x-21-x^{2}-90=0$$

Subtract 90 from -21 to get -111.

$$20x-111-x^{2}=0$$

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.

$$-x^{2}+20x-111=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 20 for b, and -111 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-1\right)\left(-111\right)}}{2\left(-1\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-1\right)\left(-111\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-20±\sqrt{400+4\left(-111\right)}}{2\left(-1\right)}$$

Multiply 4 times -111.

$$x=\frac{-20±\sqrt{400-444}}{2\left(-1\right)}$$

Add 400 to -444.

$$x=\frac{-20±\sqrt{-44}}{2\left(-1\right)}$$

Take the square root of -44.

$$x=\frac{-20±2\sqrt{11}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-20±2\sqrt{11}i}{-2}$$

Now solve the equation x=\frac{-20±2\sqrt{11}i}{-2} when ± is plus. Add -20 to 2i\sqrt{11}.

$$x=\frac{-20+2\sqrt{11}i}{-2}$$

Divide -20+2i\sqrt{11} by -2.

$$x=-\sqrt{11}i+10$$

Now solve the equation x=\frac{-20±2\sqrt{11}i}{-2} when ± is minus. Subtract 2i\sqrt{11} from -20.

$$x=\frac{-2\sqrt{11}i-20}{-2}$$

Divide -20-2i\sqrt{11} by -2.

$$x=10+\sqrt{11}i$$

The equation is now solved.

$$x=-\sqrt{11}i+10 x=10+\sqrt{11}i$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$15\left(x-1\right)-\left(x-3\right)\left(x-2\right)=90$$

Use the distributive property to multiply 15 by x-1.

$$15x-15-\left(x-3\right)\left(x-2\right)=90$$

Apply the distributive property by multiplying each term of x-3 by each term of x-2.

$$15x-15-\left(x^{2}-2x-3x+6\right)=90$$

Combine -2x and -3x to get -5x.

$$15x-15-\left(x^{2}-5x+6\right)=90$$

To find the opposite of x^{2}-5x+6, find the opposite of each term.

$$15x-15-x^{2}-\left(-5x\right)-6=90$$

The opposite of -5x is 5x.

$$15x-15-x^{2}+5x-6=90$$

Combine 15x and 5x to get 20x.

$$20x-15-x^{2}-6=90$$

Subtract 6 from -15 to get -21.

$$20x-21-x^{2}=90$$

Add 21 to both sides.

$$20x-x^{2}=90+21$$

Add 90 and 21 to get 111.

$$20x-x^{2}=111$$

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.

$$-x^{2}+20x=111$$

Divide both sides by -1.

$$\frac{-x^{2}+20x}{-1}=\frac{111}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{20}{-1}x=\frac{111}{-1}$$

Divide 20 by -1.

$$x^{2}-20x=\frac{111}{-1}$$

Divide 111 by -1.

$$x^{2}-20x=-111$$

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

$$x^{2}-20x+\left(-10\right)^{2}=-111+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-111+100$$

Add -111 to 100.

$$x^{2}-20x+100=-11$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=-11$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-10=\sqrt{11}i x-10=-\sqrt{11}i$$

Add 10 to both sides of the equation.

$$x=10+\sqrt{11}i x=-\sqrt{11}i+10$$