Solve (3-h)^2+3=2h

$$( 3 - h ) ^ { 2 } + 3 = 2 h$$

Solve for h

$$h=2$$

$$h=6$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-h\right)^{2}.

$$9-6h+h^{2}+3=2h$$

Add 9 and 3 to get 12.

$$12-6h+h^{2}=2h$$

Subtract 2h from both sides.

$$12-6h+h^{2}-2h=0$$

Combine -6h and -2h to get -8h.

$$12-8h+h^{2}=0$$

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

$$h^{2}-8h+12=0$$

To solve the equation, factor h^{2}-8h+12 using formula h^{2}+\left(a+b\right)h+ab=\left(h+a\right)\left(h+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=12$$

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 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 -8.

$$a=-6 b=-2$$

Rewrite factored expression \left(h+a\right)\left(h+b\right) using the obtained values.

$$\left(h-6\right)\left(h-2\right)$$

To find equation solutions, solve h-6=0 and h-2=0.

$$h=6 h=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-h\right)^{2}.

$$9-6h+h^{2}+3=2h$$

Add 9 and 3 to get 12.

$$12-6h+h^{2}=2h$$

Subtract 2h from both sides.

$$12-6h+h^{2}-2h=0$$

Combine -6h and -2h to get -8h.

$$12-8h+h^{2}=0$$

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

$$h^{2}-8h+12=0$$

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

$$a+b=-8 ab=1\times 12=12$$

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 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 -8.

$$a=-6 b=-2$$

Rewrite h^{2}-8h+12 as \left(h^{2}-6h\right)+\left(-2h+12\right).

$$\left(h^{2}-6h\right)+\left(-2h+12\right)$$

Factor out h in the first and -2 in the second group.

$$h\left(h-6\right)-2\left(h-6\right)$$

Factor out common term h-6 by using distributive property.

$$\left(h-6\right)\left(h-2\right)$$

To find equation solutions, solve h-6=0 and h-2=0.

$$h=6 h=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-h\right)^{2}.

$$9-6h+h^{2}+3=2h$$

Add 9 and 3 to get 12.

$$12-6h+h^{2}=2h$$

Subtract 2h from both sides.

$$12-6h+h^{2}-2h=0$$

Combine -6h and -2h to get -8h.

$$12-8h+h^{2}=0$$

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.

$$h^{2}-8h+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$h=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12}}{2}$$

Square -8.

$$h=\frac{-\left(-8\right)±\sqrt{64-4\times 12}}{2}$$

Multiply -4 times 12.

$$h=\frac{-\left(-8\right)±\sqrt{64-48}}{2}$$

Add 64 to -48.

$$h=\frac{-\left(-8\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$h=\frac{-\left(-8\right)±4}{2}$$

The opposite of -8 is 8.

$$h=\frac{8±4}{2}$$

Now solve the equation h=\frac{8±4}{2} when ± is plus. Add 8 to 4.

$$h=\frac{12}{2}$$

Divide 12 by 2.

$$h=6$$

Now solve the equation h=\frac{8±4}{2} when ± is minus. Subtract 4 from 8.

$$h=\frac{4}{2}$$

Divide 4 by 2.

$$h=2$$

The equation is now solved.

$$h=6 h=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-h\right)^{2}.

$$9-6h+h^{2}+3=2h$$

Add 9 and 3 to get 12.

$$12-6h+h^{2}=2h$$

Subtract 2h from both sides.

$$12-6h+h^{2}-2h=0$$

Combine -6h and -2h to get -8h.

$$12-8h+h^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-8h+h^{2}=-12$$

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.

$$h^{2}-8h=-12$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$h^{2}-8h+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$h^{2}-8h+16=-12+16$$

Add -12 to 16.

$$h^{2}-8h+16=4$$

Factor h^{2}-8h+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(h-4\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(h-4\right)^{2}}=\sqrt{4}$$

Simplify.

$$h-4=2 h-4=-2$$

Add 4 to both sides of the equation.

$$h=6 h=2$$