Solve (x)=-2x^2+8x+40

$$( x ) = - 2 x ^ { 2 } + 8 x + 40$$

Solve for x

$$x = \frac{3 \sqrt{41} + 7}{4} \approx 6.552343178$$

$$x=\frac{7-3\sqrt{41}}{4}\approx -3.052343178$$

Steps Using the Quadratic Formula

Add 2x^{2} to both sides.

$$x+2x^{2}=8x+40$$

Subtract 8x from both sides.

$$x+2x^{2}-8x=40$$

Combine x and -8x to get -7x.

$$-7x+2x^{2}=40$$

Subtract 40 from both sides.

$$-7x+2x^{2}-40=0$$

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.

$$2x^{2}-7x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-40\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-40\right)}}{2\times 2}$$

Multiply -8 times -40.

$$x=\frac{-\left(-7\right)±\sqrt{49+320}}{2\times 2}$$

Add 49 to 320.

$$x=\frac{-\left(-7\right)±\sqrt{369}}{2\times 2}$$

Take the square root of 369.

$$x=\frac{-\left(-7\right)±3\sqrt{41}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±3\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±3\sqrt{41}}{4}$$

Now solve the equation x=\frac{7±3\sqrt{41}}{4} when ± is plus. Add 7 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}+7}{4}$$

Now solve the equation x=\frac{7±3\sqrt{41}}{4} when ± is minus. Subtract 3\sqrt{41} from 7.

$$x=\frac{7-3\sqrt{41}}{4}$$

The equation is now solved.

$$x=\frac{3\sqrt{41}+7}{4} x=\frac{7-3\sqrt{41}}{4}$$

Steps for Completing the Square

Add 2x^{2} to both sides.

$$x+2x^{2}=8x+40$$

Subtract 8x from both sides.

$$x+2x^{2}-8x=40$$

Combine x and -8x to get -7x.

$$-7x+2x^{2}=40$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-7x=40$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{40}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{40}{2}$$

Divide 40 by 2.

$$x^{2}-\frac{7}{2}x=20$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=20+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=20+\frac{49}{16}$$

Add 20 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{369}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{369}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{369}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{3\sqrt{41}}{4} x-\frac{7}{4}=-\frac{3\sqrt{41}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{3\sqrt{41}+7}{4} x=\frac{7-3\sqrt{41}}{4}$$