Solve -2x^2+40x+100

$$-2 { x }^{ 2 } +40x+100$$

Factor

$$-2\left(x-\left(10-5\sqrt{6}\right)\right)\left(x-\left(5\sqrt{6}+10\right)\right)$$

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.

$$-2x^{2}+40x+100=0$$

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.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-2\right)\times 100}}{2\left(-2\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-2\right)\times 100}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-40±\sqrt{1600+8\times 100}}{2\left(-2\right)}$$

Multiply 8 times 100.

$$x=\frac{-40±\sqrt{1600+800}}{2\left(-2\right)}$$

Add 1600 to 800.

$$x=\frac{-40±\sqrt{2400}}{2\left(-2\right)}$$

Take the square root of 2400.

$$x=\frac{-40±20\sqrt{6}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-40±20\sqrt{6}}{-4}$$

Now solve the equation x=\frac{-40±20\sqrt{6}}{-4} when ± is plus. Add -40 to 20\sqrt{6}.

$$x=\frac{20\sqrt{6}-40}{-4}$$

Divide -40+20\sqrt{6} by -4.

$$x=10-5\sqrt{6}$$

Now solve the equation x=\frac{-40±20\sqrt{6}}{-4} when ± is minus. Subtract 20\sqrt{6} from -40.

$$x=\frac{-20\sqrt{6}-40}{-4}$$

Divide -40-20\sqrt{6} by -4.

$$x=5\sqrt{6}+10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 10-5\sqrt{6} for x_{1} and 10+5\sqrt{6} for x_{2}.

$$-2x^{2}+40x+100=-2\left(x-\left(10-5\sqrt{6}\right)\right)\left(x-\left(5\sqrt{6}+10\right)\right)$$

Evaluate

$$100+40x-2x^{2}$$