Solve 4t^2-8t-20

$$4 t ^ { 2 } - 8 t - 20$$

Factor

$$4\left(t-\left(1-\sqrt{6}\right)\right)\left(t-\left(\sqrt{6}+1\right)\right)$$

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.

$$4t^{2}-8t-20=0$$

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.

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

Square -8.

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

Multiply -4 times 4.

$$t=\frac{-\left(-8\right)±\sqrt{64-16\left(-20\right)}}{2\times 4}$$

Multiply -16 times -20.

$$t=\frac{-\left(-8\right)±\sqrt{64+320}}{2\times 4}$$

Add 64 to 320.

$$t=\frac{-\left(-8\right)±\sqrt{384}}{2\times 4}$$

Take the square root of 384.

$$t=\frac{-\left(-8\right)±8\sqrt{6}}{2\times 4}$$

The opposite of -8 is 8.

$$t=\frac{8±8\sqrt{6}}{2\times 4}$$

Multiply 2 times 4.

$$t=\frac{8±8\sqrt{6}}{8}$$

Now solve the equation t=\frac{8±8\sqrt{6}}{8} when ± is plus. Add 8 to 8\sqrt{6}.

$$t=\frac{8\sqrt{6}+8}{8}$$

Divide 8+8\sqrt{6} by 8.

$$t=\sqrt{6}+1$$

Now solve the equation t=\frac{8±8\sqrt{6}}{8} when ± is minus. Subtract 8\sqrt{6} from 8.

$$t=\frac{8-8\sqrt{6}}{8}$$

Divide 8-8\sqrt{6} by 8.

$$t=1-\sqrt{6}$$

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

$$4t^{2}-8t-20=4\left(t-\left(\sqrt{6}+1\right)\right)\left(t-\left(1-\sqrt{6}\right)\right)$$

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.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 -2x -5 = 0$$

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

$$r + s = 2 rs = -5$$

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

$$r = 1 - u s = 1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -5

$$(1 - u) (1 + u) = -5$$

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

$$1 - u^2 = -5$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -5-1 = -6$$

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

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

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

$$r =1 - \sqrt{6} = -1.449 s = 1 + \sqrt{6} = 3.449$$

Evaluate

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