Solve a^2+40a=-300

$$a ^ { 2 } + 40 a = - 300$$

Solve for a

$$a=-30$$

$$a=-10$$

Steps Using Factoring

Add 300 to both sides.

$$a^{2}+40a+300=0$$

To solve the equation, factor a^{2}+40a+300 using formula a^{2}+\left(a+b\right)a+ab=\left(a+a\right)\left(a+b\right). To find a and b, set up a system to be solved.

$$a+b=40 ab=300$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 300.

$$1,300 2,150 3,100 4,75 5,60 6,50 10,30 12,25 15,20$$

Calculate the sum for each pair.

$$1+300=301 2+150=152 3+100=103 4+75=79 5+60=65 6+50=56 10+30=40 12+25=37 15+20=35$$

The solution is the pair that gives sum 40.

$$a=10 b=30$$

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

$$\left(a+10\right)\left(a+30\right)$$

To find equation solutions, solve a+10=0 and a+30=0.

$$a=-10 a=-30$$

Steps Using Factoring By Grouping

Add 300 to both sides.

$$a^{2}+40a+300=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba+300. To find a and b, set up a system to be solved.

$$a+b=40 ab=1\times 300=300$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 300.

$$1,300 2,150 3,100 4,75 5,60 6,50 10,30 12,25 15,20$$

Calculate the sum for each pair.

$$1+300=301 2+150=152 3+100=103 4+75=79 5+60=65 6+50=56 10+30=40 12+25=37 15+20=35$$

The solution is the pair that gives sum 40.

$$a=10 b=30$$

Rewrite a^{2}+40a+300 as \left(a^{2}+10a\right)+\left(30a+300\right).

$$\left(a^{2}+10a\right)+\left(30a+300\right)$$

Factor out a in the first and 30 in the second group.

$$a\left(a+10\right)+30\left(a+10\right)$$

Factor out common term a+10 by using distributive property.

$$\left(a+10\right)\left(a+30\right)$$

To find equation solutions, solve a+10=0 and a+30=0.

$$a=-10 a=-30$$

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.

$$a^{2}+40a=-300$$

Add 300 to both sides of the equation.

$$a^{2}+40a-\left(-300\right)=-300-\left(-300\right)$$

Subtracting -300 from itself leaves 0.

$$a^{2}+40a-\left(-300\right)=0$$

Subtract -300 from 0.

$$a^{2}+40a+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 40 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-40±\sqrt{40^{2}-4\times 300}}{2}$$

Square 40.

$$a=\frac{-40±\sqrt{1600-4\times 300}}{2}$$

Multiply -4 times 300.

$$a=\frac{-40±\sqrt{1600-1200}}{2}$$

Add 1600 to -1200.

$$a=\frac{-40±\sqrt{400}}{2}$$

Take the square root of 400.

$$a=\frac{-40±20}{2}$$

Now solve the equation a=\frac{-40±20}{2} when ± is plus. Add -40 to 20.

$$a=-\frac{20}{2}$$

Divide -20 by 2.

$$a=-10$$

Now solve the equation a=\frac{-40±20}{2} when ± is minus. Subtract 20 from -40.

$$a=-\frac{60}{2}$$

Divide -60 by 2.

$$a=-30$$

The equation is now solved.

$$a=-10 a=-30$$

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.

$$a^{2}+40a=-300$$

Divide 40, the coefficient of the x term, by 2 to get 20. Then add the square of 20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+40a+20^{2}=-300+20^{2}$$

Square 20.

$$a^{2}+40a+400=-300+400$$

Add -300 to 400.

$$a^{2}+40a+400=100$$

Factor a^{2}+40a+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+20\right)^{2}=100$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+20\right)^{2}}=\sqrt{100}$$

Simplify.

$$a+20=10 a+20=-10$$

Subtract 20 from both sides of the equation.

$$a=-10 a=-30$$