Solve -x^2+2x-9=-5x+3

$$- x ^ { 2 } + 2 x - 9 = - 5 x + 3$$

Solve for x

$$x=3$$

$$x=4$$

Steps Using Factoring By Grouping

Add 5x to both sides.

$$-x^{2}+2x-9+5x=3$$

Combine 2x and 5x to get 7x.

$$-x^{2}+7x-9=3$$

Subtract 3 from both sides.

$$-x^{2}+7x-9-3=0$$

Subtract 3 from -9 to get -12.

$$-x^{2}+7x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=7 ab=-\left(-12\right)=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=4 b=3$$

Rewrite -x^{2}+7x-12 as \left(-x^{2}+4x\right)+\left(3x-12\right).

$$\left(-x^{2}+4x\right)+\left(3x-12\right)$$

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

$$-x\left(x-4\right)+3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x+3\right)$$

To find equation solutions, solve x-4=0 and -x+3=0.

$$x=4 x=3$$

Steps Using the Quadratic Formula

Add 5x to both sides.

$$-x^{2}+2x-9+5x=3$$

Combine 2x and 5x to get 7x.

$$-x^{2}+7x-9=3$$

Subtract 3 from both sides.

$$-x^{2}+7x-9-3=0$$

Subtract 3 from -9 to get -12.

$$-x^{2}+7x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 7 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-7±\sqrt{49-48}}{2\left(-1\right)}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±1}{-2}$$

Now solve the equation x=\frac{-7±1}{-2} when ± is plus. Add -7 to 1.

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

Divide -6 by -2.

$$x=3$$

Now solve the equation x=\frac{-7±1}{-2} when ± is minus. Subtract 1 from -7.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=3 x=4$$

Steps for Completing the Square

Add 5x to both sides.

$$-x^{2}+2x-9+5x=3$$

Combine 2x and 5x to get 7x.

$$-x^{2}+7x-9=3$$

Add 9 to both sides.

$$-x^{2}+7x=3+9$$

Add 3 and 9 to get 12.

$$-x^{2}+7x=12$$

Divide both sides by -1.

$$\frac{-x^{2}+7x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{7}{-1}x=\frac{12}{-1}$$

Divide 7 by -1.

$$x^{2}-7x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-7x=-12$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=4 x=3$$