Solve 3a^2-72a+540=300

$$3 a ^ { 2 } - 72 a + 540 = 300$$

Solve for a

$$a=4$$

$$a=20$$

Steps Using Factoring By Grouping

Subtract 300 from both sides.

$$3a^{2}-72a+540-300=0$$

Subtract 300 from 540 to get 240.

$$3a^{2}-72a+240=0$$

Divide both sides by 3.

$$a^{2}-24a+80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+80. To find a and b, set up a system to be solved.

$$a+b=-24 ab=1\times 80=80$$

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 80.

$$-1,-80 -2,-40 -4,-20 -5,-16 -8,-10$$

Calculate the sum for each pair.

$$-1-80=-81 -2-40=-42 -4-20=-24 -5-16=-21 -8-10=-18$$

The solution is the pair that gives sum -24.

$$a=-20 b=-4$$

Rewrite a^{2}-24a+80 as \left(a^{2}-20a\right)+\left(-4a+80\right).

$$\left(a^{2}-20a\right)+\left(-4a+80\right)$$

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

$$a\left(a-20\right)-4\left(a-20\right)$$

Factor out common term a-20 by using distributive property.

$$\left(a-20\right)\left(a-4\right)$$

To find equation solutions, solve a-20=0 and a-4=0.

$$a=20 a=4$$

Steps Using the Quadratic Formula

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.

$$3a^{2}-72a+540=300$$

Subtract 300 from both sides of the equation.

$$3a^{2}-72a+540-300=300-300$$

Subtracting 300 from itself leaves 0.

$$3a^{2}-72a+540-300=0$$

Subtract 300 from 540.

$$3a^{2}-72a+240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -72 for b, and 240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-72\right)±\sqrt{\left(-72\right)^{2}-4\times 3\times 240}}{2\times 3}$$

Square -72.

$$a=\frac{-\left(-72\right)±\sqrt{5184-4\times 3\times 240}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-72\right)±\sqrt{5184-12\times 240}}{2\times 3}$$

Multiply -12 times 240.

$$a=\frac{-\left(-72\right)±\sqrt{5184-2880}}{2\times 3}$$

Add 5184 to -2880.

$$a=\frac{-\left(-72\right)±\sqrt{2304}}{2\times 3}$$

Take the square root of 2304.

$$a=\frac{-\left(-72\right)±48}{2\times 3}$$

The opposite of -72 is 72.

$$a=\frac{72±48}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{72±48}{6}$$

Now solve the equation a=\frac{72±48}{6} when ± is plus. Add 72 to 48.

$$a=\frac{120}{6}$$

Divide 120 by 6.

$$a=20$$

Now solve the equation a=\frac{72±48}{6} when ± is minus. Subtract 48 from 72.

$$a=\frac{24}{6}$$

Divide 24 by 6.

$$a=4$$

The equation is now solved.

$$a=20 a=4$$

Steps for Completing the Square

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.

$$3a^{2}-72a+540=300$$

Subtract 540 from both sides of the equation.

$$3a^{2}-72a+540-540=300-540$$

Subtracting 540 from itself leaves 0.

$$3a^{2}-72a=300-540$$

Subtract 540 from 300.

$$3a^{2}-72a=-240$$

Divide both sides by 3.

$$\frac{3a^{2}-72a}{3}=-\frac{240}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a^{2}+\left(-\frac{72}{3}\right)a=-\frac{240}{3}$$

Divide -72 by 3.

$$a^{2}-24a=-\frac{240}{3}$$

Divide -240 by 3.

$$a^{2}-24a=-80$$

Divide -24, the coefficient of the x term, by 2 to get -12. Then add the square of -12 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-24a+\left(-12\right)^{2}=-80+\left(-12\right)^{2}$$

Square -12.

$$a^{2}-24a+144=-80+144$$

Add -80 to 144.

$$a^{2}-24a+144=64$$

Factor a^{2}-24a+144. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-12\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-12\right)^{2}}=\sqrt{64}$$

Simplify.

$$a-12=8 a-12=-8$$

Add 12 to both sides of the equation.

$$a=20 a=4$$