Solve 0=4+7x+2x^2

$$0=4+7x+2 { x }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{17}-7}{4}\approx -0.719223594$$

$$x=\frac{-\sqrt{17}-7}{4}\approx -2.780776406$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4+7x+2x^{2}=0$$

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.

$$2x^{2}+7x+4=0$$

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

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 4}}{2\times 2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 2\times 4}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-7±\sqrt{49-8\times 4}}{2\times 2}$$

Multiply -8 times 4.

$$x=\frac{-7±\sqrt{49-32}}{2\times 2}$$

Add 49 to -32.

$$x=\frac{-7±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-7±\sqrt{17}}{4}$$

Now solve the equation x=\frac{-7±\sqrt{17}}{4} when ± is plus. Add -7 to \sqrt{17}.

$$x=\frac{\sqrt{17}-7}{4}$$

Now solve the equation x=\frac{-7±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from -7.

$$x=\frac{-\sqrt{17}-7}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-7}{4} x=\frac{-\sqrt{17}-7}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4+7x+2x^{2}=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$7x+2x^{2}=-4$$

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.

$$2x^{2}+7x=-4$$

Divide both sides by 2.

$$\frac{2x^{2}+7x}{2}=-\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{7}{2}x=-\frac{4}{2}$$

Divide -4 by 2.

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

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.

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

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

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

Add -2 to \frac{49}{16}.

$$x^{2}+\frac{7}{2}x+\frac{49}{16}=\frac{17}{16}$$

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}.

$$\left(x+\frac{7}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{7}{4}=\frac{\sqrt{17}}{4} x+\frac{7}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{7}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-7}{4} x=\frac{-\sqrt{17}-7}{4}$$