Solve 2-3x(4-x)=x^2-16

$$2 - 3 x ( 4 - x ) = x ^ { 2 } - 16$$

Solve for x

$$x=3$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$2-3x\left(4-x\right)-x^{2}=-16$$

Add 16 to both sides.

$$2-3x\left(4-x\right)-x^{2}+16=0$$

Use the distributive property to multiply -3x by 4-x.

$$2-12x+3x^{2}-x^{2}+16=0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2-12x+2x^{2}+16=0$$

Add 2 and 16 to get 18.

$$18-12x+2x^{2}=0$$

Divide both sides by 2.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

$$a+b=-6 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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.

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

Rewrite as a binomial square.

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

To find equation solution, solve x-3=0.

$$x=3$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$2-3x\left(4-x\right)-x^{2}=-16$$

Add 16 to both sides.

$$2-3x\left(4-x\right)-x^{2}+16=0$$

Use the distributive property to multiply -3x by 4-x.

$$2-12x+3x^{2}-x^{2}+16=0$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2-12x+2x^{2}+16=0$$

Add 2 and 16 to get 18.

$$18-12x+2x^{2}=0$$

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.

$$2x^{2}-12x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 18}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 18}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2\times 2}$$

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{-12}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$2-3x\left(4-x\right)-x^{2}=-16$$

Use the distributive property to multiply -3x by 4-x.

$$2-12x+3x^{2}-x^{2}=-16$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$2-12x+2x^{2}=-16$$

Subtract 2 from both sides.

$$-12x+2x^{2}=-16-2$$

Subtract 2 from -16 to get -18.

$$-12x+2x^{2}=-18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-12x=-18$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=-\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=-\frac{18}{2}$$

Divide -12 by 2.

$$x^{2}-6x=-\frac{18}{2}$$

Divide -18 by 2.

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

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.

$$x=3$$