Solve 4=-x^2+6x-5

$$4 = - x ^ { 2 } + 6 x - 5$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+6x-5=4$$

Subtract 4 from both sides.

$$-x^{2}+6x-5-4=0$$

Subtract 4 from -5 to get -9.

$$-x^{2}+6x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite -x^{2}+6x-9 as \left(-x^{2}+3x\right)+\left(3x-9\right).

$$\left(-x^{2}+3x\right)+\left(3x-9\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-3\right)+3\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(-x+3\right)$$

To find equation solutions, solve x-3=0 and -x+3=0.

$$x=3 x=3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+6x-5=4$$

Subtract 4 from both sides.

$$-x^{2}+6x-5-4=0$$

Subtract 4 from -5 to get -9.

$$-x^{2}+6x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-6±\sqrt{36+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-6±\sqrt{36-36}}{2\left(-1\right)}$$

Add 36 to -36.

$$x=\frac{-6±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=-\frac{6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+6x-5=4$$

Add 5 to both sides.

$$-x^{2}+6x=4+5$$

Add 4 and 5 to get 9.

$$-x^{2}+6x=9$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{9}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{9}{-1}$$

Divide 9 by -1.

$$x^{2}-6x=-9$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$