Solve f(x)=-2x^2+x+5

$$f ( x ) = - 2 x ^ { 2 } + x + 5$$

Factor

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

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.

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

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.

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

Square 1.

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

Multiply -4 times -2.

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

Multiply 8 times 5.

$$x=\frac{-1±\sqrt{1+40}}{2\left(-2\right)}$$

Add 1 to 40.

$$x=\frac{-1±\sqrt{41}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-1±\sqrt{41}}{-4}$$

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

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

Divide -1+\sqrt{41} by -4.

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

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

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

Divide -1-\sqrt{41} by -4.

$$x=\frac{\sqrt{41}+1}{4}$$

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}.

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

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.

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

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.

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

Evaluate

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