Solve 2w^2-6w+28=(w+2)^2

$$2 w ^ { 2 } - 6 w + 28 = ( w + 2 ) ^ { 2 }$$

Solve for w

$$w=4$$

$$w=6$$

Steps Using Factoring

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

$$2w^{2}-6w+28=w^{2}+4w+4$$

Subtract w^{2} from both sides.

$$2w^{2}-6w+28-w^{2}=4w+4$$

Combine 2w^{2} and -w^{2} to get w^{2}.

$$w^{2}-6w+28=4w+4$$

Subtract 4w from both sides.

$$w^{2}-6w+28-4w=4$$

Combine -6w and -4w to get -10w.

$$w^{2}-10w+28=4$$

Subtract 4 from both sides.

$$w^{2}-10w+28-4=0$$

Subtract 4 from 28 to get 24.

$$w^{2}-10w+24=0$$

To solve the equation, factor w^{2}-10w+24 using formula w^{2}+\left(a+b\right)w+ab=\left(w+a\right)\left(w+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

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

$$\left(w-6\right)\left(w-4\right)$$

To find equation solutions, solve w-6=0 and w-4=0.

$$w=6 w=4$$

Steps Using Factoring By Grouping

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

$$2w^{2}-6w+28=w^{2}+4w+4$$

Subtract w^{2} from both sides.

$$2w^{2}-6w+28-w^{2}=4w+4$$

Combine 2w^{2} and -w^{2} to get w^{2}.

$$w^{2}-6w+28=4w+4$$

Subtract 4w from both sides.

$$w^{2}-6w+28-4w=4$$

Combine -6w and -4w to get -10w.

$$w^{2}-10w+28=4$$

Subtract 4 from both sides.

$$w^{2}-10w+28-4=0$$

Subtract 4 from 28 to get 24.

$$w^{2}-10w+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 24=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -10.

$$a=-6 b=-4$$

Rewrite w^{2}-10w+24 as \left(w^{2}-6w\right)+\left(-4w+24\right).

$$\left(w^{2}-6w\right)+\left(-4w+24\right)$$

Factor out w in the first and -4 in the second group.

$$w\left(w-6\right)-4\left(w-6\right)$$

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

$$\left(w-6\right)\left(w-4\right)$$

To find equation solutions, solve w-6=0 and w-4=0.

$$w=6 w=4$$

Steps Using the Quadratic Formula

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

$$2w^{2}-6w+28=w^{2}+4w+4$$

Subtract w^{2} from both sides.

$$2w^{2}-6w+28-w^{2}=4w+4$$

Combine 2w^{2} and -w^{2} to get w^{2}.

$$w^{2}-6w+28=4w+4$$

Subtract 4w from both sides.

$$w^{2}-6w+28-4w=4$$

Combine -6w and -4w to get -10w.

$$w^{2}-10w+28=4$$

Subtract 4 from both sides.

$$w^{2}-10w+28-4=0$$

Subtract 4 from 28 to get 24.

$$w^{2}-10w+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 24}}{2}$$

Square -10.

$$w=\frac{-\left(-10\right)±\sqrt{100-4\times 24}}{2}$$

Multiply -4 times 24.

$$w=\frac{-\left(-10\right)±\sqrt{100-96}}{2}$$

Add 100 to -96.

$$w=\frac{-\left(-10\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$w=\frac{-\left(-10\right)±2}{2}$$

The opposite of -10 is 10.

$$w=\frac{10±2}{2}$$

Now solve the equation w=\frac{10±2}{2} when ± is plus. Add 10 to 2.

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

Divide 12 by 2.

$$w=6$$

Now solve the equation w=\frac{10±2}{2} when ± is minus. Subtract 2 from 10.

$$w=\frac{8}{2}$$

Divide 8 by 2.

$$w=4$$

The equation is now solved.

$$w=6 w=4$$

Steps for Completing the Square

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

$$2w^{2}-6w+28=w^{2}+4w+4$$

Subtract w^{2} from both sides.

$$2w^{2}-6w+28-w^{2}=4w+4$$

Combine 2w^{2} and -w^{2} to get w^{2}.

$$w^{2}-6w+28=4w+4$$

Subtract 4w from both sides.

$$w^{2}-6w+28-4w=4$$

Combine -6w and -4w to get -10w.

$$w^{2}-10w+28=4$$

Subtract 28 from both sides.

$$w^{2}-10w=4-28$$

Subtract 28 from 4 to get -24.

$$w^{2}-10w=-24$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}-10w+\left(-5\right)^{2}=-24+\left(-5\right)^{2}$$

Square -5.

$$w^{2}-10w+25=-24+25$$

Add -24 to 25.

$$w^{2}-10w+25=1$$

Factor w^{2}-10w+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(w-5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w-5\right)^{2}}=\sqrt{1}$$

Simplify.

$$w-5=1 w-5=-1$$

Add 5 to both sides of the equation.

$$w=6 w=4$$