Solve f(t)=-2t^2+120t-1600

$$f ( t ) = - 2 t ^ { 2 } + 120 t - 1600$$

Factor

$$2\left(20-t\right)\left(t-40\right)$$

Solution Steps

Factor out 2.

$$2\left(-t^{2}+60t-800\right)$$

Consider -t^{2}+60t-800. Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt-800. To find a and b, set up a system to be solved.

$$a+b=60 ab=-\left(-800\right)=800$$

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 800.

$$1,800 2,400 4,200 5,160 8,100 10,80 16,50 20,40 25,32$$

Calculate the sum for each pair.

$$1+800=801 2+400=402 4+200=204 5+160=165 8+100=108 10+80=90 16+50=66 20+40=60 25+32=57$$

The solution is the pair that gives sum 60.

$$a=40 b=20$$

Rewrite -t^{2}+60t-800 as \left(-t^{2}+40t\right)+\left(20t-800\right).

$$\left(-t^{2}+40t\right)+\left(20t-800\right)$$

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

$$-t\left(t-40\right)+20\left(t-40\right)$$

Factor out common term t-40 by using distributive property.

$$\left(t-40\right)\left(-t+20\right)$$

Rewrite the complete factored expression.

$$2\left(t-40\right)\left(-t+20\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-2t^{2}+120t-1600=0$$

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.

$$t=\frac{-120±\sqrt{120^{2}-4\left(-2\right)\left(-1600\right)}}{2\left(-2\right)}$$

Square 120.

$$t=\frac{-120±\sqrt{14400-4\left(-2\right)\left(-1600\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$t=\frac{-120±\sqrt{14400+8\left(-1600\right)}}{2\left(-2\right)}$$

Multiply 8 times -1600.

$$t=\frac{-120±\sqrt{14400-12800}}{2\left(-2\right)}$$

Add 14400 to -12800.

$$t=\frac{-120±\sqrt{1600}}{2\left(-2\right)}$$

Take the square root of 1600.

$$t=\frac{-120±40}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{-120±40}{-4}$$

Now solve the equation t=\frac{-120±40}{-4} when ± is plus. Add -120 to 40.

$$t=-\frac{80}{-4}$$

Divide -80 by -4.

$$t=20$$

Now solve the equation t=\frac{-120±40}{-4} when ± is minus. Subtract 40 from -120.

$$t=-\frac{160}{-4}$$

Divide -160 by -4.

$$t=40$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and 40 for x_{2}.

$$-2t^{2}+120t-1600=-2\left(t-20\right)\left(t-40\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 -60x +800 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 60 rs = 800$$

Two numbers r and s sum up to 60 exactly when the average of the two numbers is \frac{1}{2}*60 = 30. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 30 - u s = 30 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 800

$$(30 - u) (30 + u) = 800$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$900 - u^2 = 800$$

Simplify the expression by subtracting 900 on both sides

$$-u^2 = 800-900 = -100$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 100 u = \pm\sqrt{100} = \pm 10$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =30 - 10 = 20 s = 30 + 10 = 40$$

Evaluate

$$2\left(20-t\right)\left(t-40\right)$$