Solve -3x^2-33x+45=9x

$$- 3 x ^ { 2 } - 33 x + 45 = 9 x$$

Solve for x

$$x=-15$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 9x from both sides.

$$-3x^{2}-33x+45-9x=0$$

Combine -33x and -9x to get -42x.

$$-3x^{2}-42x+45=0$$

Divide both sides by 3.

$$-x^{2}-14x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-14 ab=-15=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -14.

$$a=1 b=-15$$

Rewrite -x^{2}-14x+15 as \left(-x^{2}+x\right)+\left(-15x+15\right).

$$\left(-x^{2}+x\right)+\left(-15x+15\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(-x+1\right)+15\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+15\right)$$

To find equation solutions, solve -x+1=0 and x+15=0.

$$x=1 x=-15$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

$$-3x^{2}-33x+45-9x=0$$

Combine -33x and -9x to get -42x.

$$-3x^{2}-42x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -42 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-42\right)±\sqrt{\left(-42\right)^{2}-4\left(-3\right)\times 45}}{2\left(-3\right)}$$

Square -42.

$$x=\frac{-\left(-42\right)±\sqrt{1764-4\left(-3\right)\times 45}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-42\right)±\sqrt{1764+12\times 45}}{2\left(-3\right)}$$

Multiply 12 times 45.

$$x=\frac{-\left(-42\right)±\sqrt{1764+540}}{2\left(-3\right)}$$

Add 1764 to 540.

$$x=\frac{-\left(-42\right)±\sqrt{2304}}{2\left(-3\right)}$$

Take the square root of 2304.

$$x=\frac{-\left(-42\right)±48}{2\left(-3\right)}$$

The opposite of -42 is 42.

$$x=\frac{42±48}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{42±48}{-6}$$

Now solve the equation x=\frac{42±48}{-6} when ± is plus. Add 42 to 48.

$$x=\frac{90}{-6}$$

Divide 90 by -6.

$$x=-15$$

Now solve the equation x=\frac{42±48}{-6} when ± is minus. Subtract 48 from 42.

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

Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=-15 x=1$$

Steps for Completing the Square

Subtract 9x from both sides.

$$-3x^{2}-33x+45-9x=0$$

Combine -33x and -9x to get -42x.

$$-3x^{2}-42x+45=0$$

Subtract 45 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}-42x=-45$$

Divide both sides by -3.

$$\frac{-3x^{2}-42x}{-3}=-\frac{45}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{42}{-3}\right)x=-\frac{45}{-3}$$

Divide -42 by -3.

$$x^{2}+14x=-\frac{45}{-3}$$

Divide -45 by -3.

$$x^{2}+14x=15$$

Divide 14, the coefficient of the x term, by 2 to get 7. Then add the square of 7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=15+7^{2}$$

Square 7.

$$x^{2}+14x+49=15+49$$

Add 15 to 49.

$$x^{2}+14x+49=64$$

Factor x^{2}+14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+7\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{64}$$

Simplify.

$$x+7=8 x+7=-8$$

Subtract 7 from both sides of the equation.

$$x=1 x=-15$$