Solve 0=3w^2-60w+300

$$0 = 3 w ^ { 2 } - 60 w + 300$$

Solve for w

$$w=10$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$3w^{2}-60w+300=0$$

Divide both sides by 3.

$$w^{2}-20w+100=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as w^{2}+aw+bw+100. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 100=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$a=-10 b=-10$$

Rewrite w^{2}-20w+100 as \left(w^{2}-10w\right)+\left(-10w+100\right).

$$\left(w^{2}-10w\right)+\left(-10w+100\right)$$

Factor out w in the first and -10 in the second group.

$$w\left(w-10\right)-10\left(w-10\right)$$

Factor out common term w-10 by using distributive property.

$$\left(w-10\right)\left(w-10\right)$$

Rewrite as a binomial square.

$$\left(w-10\right)^{2}$$

To find equation solution, solve w-10=0.

$$w=10$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3w^{2}-60w+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -60 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$w=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 3\times 300}}{2\times 3}$$

Square -60.

$$w=\frac{-\left(-60\right)±\sqrt{3600-4\times 3\times 300}}{2\times 3}$$

Multiply -4 times 3.

$$w=\frac{-\left(-60\right)±\sqrt{3600-12\times 300}}{2\times 3}$$

Multiply -12 times 300.

$$w=\frac{-\left(-60\right)±\sqrt{3600-3600}}{2\times 3}$$

Add 3600 to -3600.

$$w=\frac{-\left(-60\right)±\sqrt{0}}{2\times 3}$$

Take the square root of 0.

$$w=-\frac{-60}{2\times 3}$$

The opposite of -60 is 60.

$$w=\frac{60}{2\times 3}$$

Multiply 2 times 3.

$$w=\frac{60}{6}$$

Divide 60 by 6.

$$w=10$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3w^{2}-60w+300=0$$

Subtract 300 from both sides. Anything subtracted from zero gives its negation.

$$3w^{2}-60w=-300$$

Divide both sides by 3.

$$\frac{3w^{2}-60w}{3}=-\frac{300}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$w^{2}+\left(-\frac{60}{3}\right)w=-\frac{300}{3}$$

Divide -60 by 3.

$$w^{2}-20w=-\frac{300}{3}$$

Divide -300 by 3.

$$w^{2}-20w=-100$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$w^{2}-20w+\left(-10\right)^{2}=-100+\left(-10\right)^{2}$$

Square -10.

$$w^{2}-20w+100=-100+100$$

Add -100 to 100.

$$w^{2}-20w+100=0$$

Factor w^{2}-20w+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(w-10\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(w-10\right)^{2}}=\sqrt{0}$$

Simplify.

$$w-10=0 w-10=0$$

Add 10 to both sides of the equation.

$$w=10 w=10$$

The equation is now solved. Solutions are the same.

$$w=10$$