Solve t^2+8t-20

$$t ^ { 2 } + 8 t - 20$$

Factor

$$\left(t-2\right)\left(t+10\right)$$

Solution Steps

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.

$$a+b=8 ab=1\left(-20\right)=-20$$

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 8.

$$a=-2 b=10$$

Rewrite t^{2}+8t-20 as \left(t^{2}-2t\right)+\left(10t-20\right).

$$\left(t^{2}-2t\right)+\left(10t-20\right)$$

Factor out t in the first and 10 in the second group.

$$t\left(t-2\right)+10\left(t-2\right)$$

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(t+10\right)$$

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.

$$t^{2}+8t-20=0$$

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.

$$t=\frac{-8±\sqrt{8^{2}-4\left(-20\right)}}{2}$$

Square 8.

$$t=\frac{-8±\sqrt{64-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$t=\frac{-8±\sqrt{64+80}}{2}$$

Add 64 to 80.

$$t=\frac{-8±\sqrt{144}}{2}$$

Take the square root of 144.

$$t=\frac{-8±12}{2}$$

Now solve the equation t=\frac{-8±12}{2} when ± is plus. Add -8 to 12.

$$t=\frac{4}{2}$$

Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{-8±12}{2} when ± is minus. Subtract 12 from -8.

$$t=-\frac{20}{2}$$

Divide -20 by 2.

$$t=-10$$

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

$$t^{2}+8t-20=\left(t-2\right)\left(t-\left(-10\right)\right)$$

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

$$t^{2}+8t-20=\left(t-2\right)\left(t+10\right)$$

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.

$$x ^ 2 +8x -20 = 0$$

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

$$r + s = -8 rs = -20$$

Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>

$$r = -4 - u s = -4 + u$$

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

$$(-4 - u) (-4 + u) = -20$$

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

$$16 - u^2 = -20$$

Simplify the expression by subtracting 16 on both sides

$$-u^2 = -20-16 = -36$$

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

$$u^2 = 36 u = \pm\sqrt{36} = \pm 6$$

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

$$r =-4 - 6 = -10 s = -4 + 6 = 2$$

Evaluate

$$\left(t-2\right)\left(t+10\right)$$