Solve 2t^2-2t+6

$$2 t ^ { 2 } - 2 t + 6$$

Factor

$$2\left(t^{2}-t+3\right)$$

Solution Steps

Factor out 2. Polynomial t^{2}-t+3 is not factored since it does not have any rational roots.

$$2\left(t^{2}-t+3\right)$$

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.

$$2t^{2}-2t+6=0$$

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.

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

Square -2.

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

Multiply -4 times 2.

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

Multiply -8 times 6.

$$t=\frac{-\left(-2\right)±\sqrt{4-48}}{2\times 2}$$

Add 4 to -48.

$$t=\frac{-\left(-2\right)±\sqrt{-44}}{2\times 2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$2t^{2}-2t+6$$

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 2

$$x ^ 2 -1x +3 = 0$$

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

$$r + s = 1 rs = 3$$

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 = 3

$$(\frac{1}{2} - u) (\frac{1}{2} + u) = 3$$

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

$$\frac{1}{4} - u^2 = 3$$

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

$$-u^2 = 3-\frac{1}{4} = \frac{11}{4}$$

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

$$u^2 = -\frac{11}{4} u = \pm\sqrt{-\frac{11}{4}} = \pm \frac{\sqrt{11}}{2}i$$

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

$$r =\frac{1}{2} - \frac{\sqrt{11}}{2}i = 0.500 - 1.658i s = \frac{1}{2} + \frac{\sqrt{11}}{2}i = 0.500 + 1.658i$$

Evaluate

$$2\left(t^{2}-t+3\right)$$