Solve a^2+24a=-80

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

Solve for a

$$a=-20$$

$$a=-4$$

Steps Using Factoring

Add 80 to both sides.

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

To solve the equation, factor a^{2}+24a+80 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.

$$a+b=24 ab=80$$

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 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.

$$a=4 b=20$$

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

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

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

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

Steps Using Factoring By Grouping

Add 80 to both sides.

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

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 positive, a and b are both positive. 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.

$$a=4 b=20$$

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

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

Factor out a in the first and 20 in the second group.

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

Factor out common term a+4 by using distributive property.

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

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

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

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.

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

Add 80 to both sides of the equation.

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

Subtracting -80 from itself leaves 0.

$$a^{2}+24a-\left(-80\right)=0$$

Subtract -80 from 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 24 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-24±\sqrt{24^{2}-4\times 80}}{2}$$

Square 24.

$$a=\frac{-24±\sqrt{576-4\times 80}}{2}$$

Multiply -4 times 80.

$$a=\frac{-24±\sqrt{576-320}}{2}$$

Add 576 to -320.

$$a=\frac{-24±\sqrt{256}}{2}$$

Take the square root of 256.

$$a=\frac{-24±16}{2}$$

Now solve the equation a=\frac{-24±16}{2} when ± is plus. Add -24 to 16.

$$a=-\frac{8}{2}$$

Divide -8 by 2.

$$a=-4$$

Now solve the equation a=\frac{-24±16}{2} when ± is minus. Subtract 16 from -24.

$$a=-\frac{40}{2}$$

Divide -40 by 2.

$$a=-20$$

The equation is now solved.

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

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.

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

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.

$$a^{2}+24a+12^{2}=-80+12^{2}$$

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.

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

Subtract 12 from both sides of the equation.

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