Solve -x^2+4x+6

$$- x ^ { 2 } + 4 x + 6$$

Factor

$$-\left(x-\left(2-\sqrt{10}\right)\right)\left(x-\left(\sqrt{10}+2\right)\right)$$

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.

$$-x^{2}+4x+6=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 6}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 6}}{2\left(-1\right)}$$

Multiply 4 times 6.

$$x=\frac{-4±\sqrt{16+24}}{2\left(-1\right)}$$

Add 16 to 24.

$$x=\frac{-4±\sqrt{40}}{2\left(-1\right)}$$

Take the square root of 40.

$$x=\frac{-4±2\sqrt{10}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2\sqrt{10}}{-2}$$

Now solve the equation x=\frac{-4±2\sqrt{10}}{-2} when ± is plus. Add -4 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-4}{-2}$$

Divide -4+2\sqrt{10} by -2.

$$x=2-\sqrt{10}$$

Now solve the equation x=\frac{-4±2\sqrt{10}}{-2} when ± is minus. Subtract 2\sqrt{10} from -4.

$$x=\frac{-2\sqrt{10}-4}{-2}$$

Divide -4-2\sqrt{10} by -2.

$$x=\sqrt{10}+2$$

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

$$-x^{2}+4x+6=-\left(x-\left(2-\sqrt{10}\right)\right)\left(x-\left(\sqrt{10}+2\right)\right)$$

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.

$$x ^ 2 -4x -6 = 0$$

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

$$r + s = 4 rs = -6$$

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

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -6

$$(2 - u) (2 + u) = -6$$

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

$$4 - u^2 = -6$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -6-4 = -10$$

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

$$u^2 = 10 u = \pm\sqrt{10} = \pm \sqrt{10}$$

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

$$r =2 - \sqrt{10} = -1.162 s = 2 + \sqrt{10} = 5.162$$

Evaluate

$$6+4x-x^{2}$$