Solve 400=40t-t^2

$$400 = 40 t - t ^ { 2 }$$

Solve for t

$$t=20$$

Steps Using Factoring By Grouping

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

$$40t-t^{2}=400$$

Subtract 400 from both sides.

$$40t-t^{2}-400=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-t^{2}+40t-400=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt-400. To find a and b, set up a system to be solved.

$$a+b=40 ab=-\left(-400\right)=400$$

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 400.

$$1,400 2,200 4,100 5,80 8,50 10,40 16,25 20,20$$

Calculate the sum for each pair.

$$1+400=401 2+200=202 4+100=104 5+80=85 8+50=58 10+40=50 16+25=41 20+20=40$$

The solution is the pair that gives sum 40.

$$a=20 b=20$$

Rewrite -t^{2}+40t-400 as \left(-t^{2}+20t\right)+\left(20t-400\right).

$$\left(-t^{2}+20t\right)+\left(20t-400\right)$$

Factor out -t in the first and 20 in the second group.

$$-t\left(t-20\right)+20\left(t-20\right)$$

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

$$\left(t-20\right)\left(-t+20\right)$$

To find equation solutions, solve t-20=0 and -t+20=0.

$$t=20 t=20$$

Steps Using the Quadratic Formula

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

$$40t-t^{2}=400$$

Subtract 400 from both sides.

$$40t-t^{2}-400=0$$

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.

$$-t^{2}+40t-400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 40 for b, and -400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-40±\sqrt{40^{2}-4\left(-1\right)\left(-400\right)}}{2\left(-1\right)}$$

Square 40.

$$t=\frac{-40±\sqrt{1600-4\left(-1\right)\left(-400\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-40±\sqrt{1600+4\left(-400\right)}}{2\left(-1\right)}$$

Multiply 4 times -400.

$$t=\frac{-40±\sqrt{1600-1600}}{2\left(-1\right)}$$

Add 1600 to -1600.

$$t=\frac{-40±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$t=-\frac{40}{2\left(-1\right)}$$

Multiply 2 times -1.

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

Divide -40 by -2.

$$t=20$$

Steps for Completing the Square

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

$$40t-t^{2}=400$$

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.

$$-t^{2}+40t=400$$

Divide both sides by -1.

$$\frac{-t^{2}+40t}{-1}=\frac{400}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\frac{40}{-1}t=\frac{400}{-1}$$

Divide 40 by -1.

$$t^{2}-40t=\frac{400}{-1}$$

Divide 400 by -1.

$$t^{2}-40t=-400$$

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.

$$t^{2}-40t+\left(-20\right)^{2}=-400+\left(-20\right)^{2}$$

Square -20.

$$t^{2}-40t+400=-400+400$$

Add -400 to 400.

$$t^{2}-40t+400=0$$

Factor t^{2}-40t+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}.

$$\left(t-20\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-20\right)^{2}}=\sqrt{0}$$

Simplify.

$$t-20=0 t-20=0$$

Add 20 to both sides of the equation.

$$t=20 t=20$$

The equation is now solved. Solutions are the same.

$$t=20$$