Solve 2x^2=-35x-1

$$2 x ^ { 2 } = - 35 x - 1$$

Solve for x

$$x=\frac{\sqrt{1217}-35}{4}\approx -0.028618229$$

$$x=\frac{-\sqrt{1217}-35}{4}\approx -17.471381771$$

Steps Using the Quadratic Formula

Add 35x to both sides.

$$2x^{2}+35x=-1$$

Add 1 to both sides.

$$2x^{2}+35x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 35 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-35±\sqrt{35^{2}-4\times 2}}{2\times 2}$$

Square 35.

$$x=\frac{-35±\sqrt{1225-4\times 2}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-35±\sqrt{1225-8}}{2\times 2}$$

Add 1225 to -8.

$$x=\frac{-35±\sqrt{1217}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-35±\sqrt{1217}}{4}$$

Now solve the equation x=\frac{-35±\sqrt{1217}}{4} when ± is plus. Add -35 to \sqrt{1217}.

$$x=\frac{\sqrt{1217}-35}{4}$$

Now solve the equation x=\frac{-35±\sqrt{1217}}{4} when ± is minus. Subtract \sqrt{1217} from -35.

$$x=\frac{-\sqrt{1217}-35}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{1217}-35}{4} x=\frac{-\sqrt{1217}-35}{4}$$

Steps for Completing the Square

Add 35x to both sides.

$$2x^{2}+35x=-1$$

Divide both sides by 2.

$$\frac{2x^{2}+35x}{2}=-\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{35}{2}x=-\frac{1}{2}$$

Divide \frac{35}{2}, the coefficient of the x term, by 2 to get \frac{35}{4}. Then add the square of \frac{35}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{35}{2}x+\left(\frac{35}{4}\right)^{2}=-\frac{1}{2}+\left(\frac{35}{4}\right)^{2}$$

Square \frac{35}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=-\frac{1}{2}+\frac{1225}{16}$$

Add -\frac{1}{2} to \frac{1225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{35}{2}x+\frac{1225}{16}=\frac{1217}{16}$$

Factor x^{2}+\frac{35}{2}x+\frac{1225}{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}.

$$\left(x+\frac{35}{4}\right)^{2}=\frac{1217}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{35}{4}\right)^{2}}=\sqrt{\frac{1217}{16}}$$

Simplify.

$$x+\frac{35}{4}=\frac{\sqrt{1217}}{4} x+\frac{35}{4}=-\frac{\sqrt{1217}}{4}$$

Subtract \frac{35}{4} from both sides of the equation.

$$x=\frac{\sqrt{1217}-35}{4} x=\frac{-\sqrt{1217}-35}{4}$$