Solve 1-x(x-3)=4x-1

$$1-x(x-3)=4x-1$$

Solve for x

$$x=-2$$

$$x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-3.

$$1-\left(x^{2}-3x\right)=4x-1$$

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

$$1-x^{2}-\left(-3x\right)=4x-1$$

The opposite of -3x is 3x.

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

Subtract 4x from both sides.

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

Combine 3x and -4x to get -x.

$$1-x^{2}-x=-1$$

Add 1 to both sides.

$$1-x^{2}-x+1=0$$

Add 1 and 1 to get 2.

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

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.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times 2.

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

Add 1 to 8.

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

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±3}{-2}$$

Now solve the equation x=\frac{1±3}{-2} when ± is plus. Add 1 to 3.

$$x=\frac{4}{-2}$$

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{1±3}{-2} when ± is minus. Subtract 3 from 1.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-2 x=1$$

Steps for Completing the Square

Use the distributive property to multiply x by x-3.

$$1-\left(x^{2}-3x\right)=4x-1$$

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

$$1-x^{2}-\left(-3x\right)=4x-1$$

The opposite of -3x is 3x.

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

Subtract 4x from both sides.

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

Combine 3x and -4x to get -x.

$$1-x^{2}-x=-1$$

Subtract 1 from both sides.

$$-x^{2}-x=-1-1$$

Subtract 1 from -1 to get -2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -2 by -1.

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

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.

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

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

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

Add 2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=1 x=-2$$