Solve -2t^2+40t+100

$$- 2 t ^ { 2 } + 40 t + 100$$

Factor

$$-2\left(t-\left(10-5\sqrt{6}\right)\right)\left(t-\left(5\sqrt{6}+10\right)\right)$$

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.

$$-2t^{2}+40t+100=0$$

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.

$$t=\frac{-40±\sqrt{40^{2}-4\left(-2\right)\times 100}}{2\left(-2\right)}$$

Square 40.

$$t=\frac{-40±\sqrt{1600-4\left(-2\right)\times 100}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$t=\frac{-40±\sqrt{1600+8\times 100}}{2\left(-2\right)}$$

Multiply 8 times 100.

$$t=\frac{-40±\sqrt{1600+800}}{2\left(-2\right)}$$

Add 1600 to 800.

$$t=\frac{-40±\sqrt{2400}}{2\left(-2\right)}$$

Take the square root of 2400.

$$t=\frac{-40±20\sqrt{6}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{-40±20\sqrt{6}}{-4}$$

Now solve the equation t=\frac{-40±20\sqrt{6}}{-4} when ± is plus. Add -40 to 20\sqrt{6}.

$$t=\frac{20\sqrt{6}-40}{-4}$$

Divide -40+20\sqrt{6} by -4.

$$t=10-5\sqrt{6}$$

Now solve the equation t=\frac{-40±20\sqrt{6}}{-4} when ± is minus. Subtract 20\sqrt{6} from -40.

$$t=\frac{-20\sqrt{6}-40}{-4}$$

Divide -40-20\sqrt{6} by -4.

$$t=5\sqrt{6}+10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10-5\sqrt{6} for x_{1} and 10+5\sqrt{6} for x_{2}.

$$-2t^{2}+40t+100=-2\left(t-\left(10-5\sqrt{6}\right)\right)\left(t-\left(5\sqrt{6}+10\right)\right)$$

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.

$$x ^ 2 -20x -50 = 0$$

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

$$r + s = 20 rs = -50$$

Two numbers r and s sum up to 20 exactly when the average of the two numbers is \frac{1}{2}*20 = 10. 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>

$$r = 10 - u s = 10 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -50

$$(10 - u) (10 + u) = -50$$

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

$$100 - u^2 = -50$$

Simplify the expression by subtracting 100 on both sides

$$-u^2 = -50-100 = -150$$

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

$$u^2 = 150 u = \pm\sqrt{150} = \pm \sqrt{150}$$

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

$$r =10 - \sqrt{150} = -2.247 s = 10 + \sqrt{150} = 22.247$$

Evaluate

$$100+40t-2t^{2}$$