Solve 2t^2+t-5

$$2 t ^ { 2 } + t - 5$$

Factor

$$2\left(t-\frac{-\sqrt{41}-1}{4}\right)\left(t-\frac{\sqrt{41}-1}{4}\right)$$

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.

$$2t^{2}+t-5=0$$

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.

$$t=\frac{-1±\sqrt{1^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-1±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$t=\frac{-1±\sqrt{1+40}}{2\times 2}$$

Add 1 to 40.

$$t=\frac{-1±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-1±\sqrt{41}}{4}$$

Now solve the equation t=\frac{-1±\sqrt{41}}{4} when ± is plus. Add -1 to \sqrt{41}.

$$t=\frac{\sqrt{41}-1}{4}$$

Now solve the equation t=\frac{-1±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -1.

$$t=\frac{-\sqrt{41}-1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{41}}{4} for x_{1} and \frac{-1-\sqrt{41}}{4} for x_{2}.

$$2t^{2}+t-5=2\left(t-\frac{\sqrt{41}-1}{4}\right)\left(t-\frac{-\sqrt{41}-1}{4}\right)$$

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 2

$$x ^ 2 +\frac{1}{2}x -\frac{5}{2} = 0$$

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

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

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

$$r = -\frac{1}{4} - u s = -\frac{1}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{2}

$$(-\frac{1}{4} - u) (-\frac{1}{4} + u) = -\frac{5}{2}$$

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

$$\frac{1}{16} - u^2 = -\frac{5}{2}$$

Simplify the expression by subtracting \frac{1}{16} on both sides

$$-u^2 = -\frac{5}{2}-\frac{1}{16} = -\frac{41}{16}$$

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

$$u^2 = \frac{41}{16} u = \pm\sqrt{\frac{41}{16}} = \pm \frac{\sqrt{41}}{4}$$

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

$$r =-\frac{1}{4} - \frac{\sqrt{41}}{4} = -1.851 s = -\frac{1}{4} + \frac{\sqrt{41}}{4} = 1.351$$

Evaluate

$$2t^{2}+t-5$$