Solve f(x)=-2x^2+20x-9

$$f ( x ) = - 2 x ^ { 2 } + 20 x - 9$$

Factor

$$-2\left(x-\left(-\frac{\sqrt{82}}{2}+5\right)\right)\left(x-\left(\frac{\sqrt{82}}{2}+5\right)\right)$$

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.

$$-2x^{2}+20x-9=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-2\right)\left(-9\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-20±\sqrt{400+8\left(-9\right)}}{2\left(-2\right)}$$

Multiply 8 times -9.

$$x=\frac{-20±\sqrt{400-72}}{2\left(-2\right)}$$

Add 400 to -72.

$$x=\frac{-20±\sqrt{328}}{2\left(-2\right)}$$

Take the square root of 328.

$$x=\frac{-20±2\sqrt{82}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-20±2\sqrt{82}}{-4}$$

Now solve the equation x=\frac{-20±2\sqrt{82}}{-4} when ± is plus. Add -20 to 2\sqrt{82}.

$$x=\frac{2\sqrt{82}-20}{-4}$$

Divide -20+2\sqrt{82} by -4.

$$x=-\frac{\sqrt{82}}{2}+5$$

Now solve the equation x=\frac{-20±2\sqrt{82}}{-4} when ± is minus. Subtract 2\sqrt{82} from -20.

$$x=\frac{-2\sqrt{82}-20}{-4}$$

Divide -20-2\sqrt{82} by -4.

$$x=\frac{\sqrt{82}}{2}+5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5-\frac{\sqrt{82}}{2} for x_{1} and 5+\frac{\sqrt{82}}{2} for x_{2}.

$$-2x^{2}+20x-9=-2\left(x-\left(-\frac{\sqrt{82}}{2}+5\right)\right)\left(x-\left(\frac{\sqrt{82}}{2}+5\right)\right)$$

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.

$$x ^ 2 -10x +\frac{9}{2} = 0$$

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

$$r + s = 10 rs = \frac{9}{2}$$

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 5 - u s = 5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{9}{2}

$$(5 - u) (5 + u) = \frac{9}{2}$$

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

$$25 - u^2 = \frac{9}{2}$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = \frac{9}{2}-25 = -\frac{41}{2}$$

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

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

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

$$r =5 - \frac{\sqrt{41}}{\sqrt{2}} = 0.472 s = 5 + \frac{\sqrt{41}}{\sqrt{2}} = 9.528$$

Evaluate

$$-2x^{2}+20x-9$$