Solve -x^2+x=3x+1

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

Subtract 3x from both sides.

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

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

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

Subtract 1 from both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-\left(-1\right)=1$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-1$$

Rewrite -x^{2}-2x-1 as \left(-x^{2}-x\right)+\left(-x-1\right).

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

Factor out x in -x^{2}-x.

$$x\left(-x-1\right)-x-1$$

Factor out common term -x-1 by using distributive property.

$$\left(-x-1\right)\left(x+1\right)$$

To find equation solutions, solve -x-1=0 and x+1=0.

$$x=-1 x=-1$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

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

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

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

Subtract 1 from both sides.

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

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

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

Square -2.

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

Multiply -4 times -1.

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

Multiply 4 times -1.

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

Add 4 to -4.

$$x=\frac{-\left(-2\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

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

The opposite of -2 is 2.

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

Multiply 2 times -1.

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

Divide 2 by -2.

$$x=-1$$

Steps for Completing the Square

Subtract 3x from both sides.

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

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

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

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

Divide 1 by -1.

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

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

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

Square 1.

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

Add -1 to 1.

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

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

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

$$x=-1 x=-1$$

The equation is now solved. Solutions are the same.

$$x=-1$$