Solve -2t^2+2t+40

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

Factor

$$-2\left(t-5\right)\left(t+4\right)$$

Solution Steps

Factor out 2.

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

Consider -t^{2}+t+20. Factor the expression by grouping. First, the expression needs to be rewritten as -t^{2}+at+bt+20. To find a and b, set up a system to be solved.

$$a+b=1 ab=-20=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=5 b=-4$$

Rewrite -t^{2}+t+20 as \left(-t^{2}+5t\right)+\left(-4t+20\right).

$$\left(-t^{2}+5t\right)+\left(-4t+20\right)$$

Factor out -t in the first and -4 in the second group.

$$-t\left(t-5\right)-4\left(t-5\right)$$

Factor out common term t-5 by using distributive property.

$$\left(t-5\right)\left(-t-4\right)$$

Rewrite the complete factored expression.

$$2\left(t-5\right)\left(-t-4\right)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times -2.

$$t=\frac{-2±\sqrt{4+8\times 40}}{2\left(-2\right)}$$

Multiply 8 times 40.

$$t=\frac{-2±\sqrt{4+320}}{2\left(-2\right)}$$

Add 4 to 320.

$$t=\frac{-2±\sqrt{324}}{2\left(-2\right)}$$

Take the square root of 324.

$$t=\frac{-2±18}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{-2±18}{-4}$$

Now solve the equation t=\frac{-2±18}{-4} when ± is plus. Add -2 to 18.

$$t=\frac{16}{-4}$$

Divide 16 by -4.

$$t=-4$$

Now solve the equation t=\frac{-2±18}{-4} when ± is minus. Subtract 18 from -2.

$$t=-\frac{20}{-4}$$

Divide -20 by -4.

$$t=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -4 for x_{1} and 5 for x_{2}.

$$-2t^{2}+2t+40=-2\left(t-\left(-4\right)\right)\left(t-5\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-2t^{2}+2t+40=-2\left(t+4\right)\left(t-5\right)$$

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.

$$x ^ 2 -1x -20 = 0$$

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

$$r + s = 1 rs = -20$$

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

$$r = \frac{1}{2} - u s = \frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -20

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = -20$$

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

$$\frac{1}{4} - u^2 = -20$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -20-\frac{1}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =\frac{1}{2} - \frac{9}{2} = -4 s = \frac{1}{2} + \frac{9}{2} = 5$$

Evaluate

$$-2\left(t-5\right)\left(t+4\right)$$