Solve y+4=-4y^2

$$y + 4 = - 4 y ^ { 2 }$$

Solve for y

$$y=\frac{-1+3\sqrt{7}i}{8}\approx -0.125+0.992156742i$$

$$y=\frac{-3\sqrt{7}i-1}{8}\approx -0.125-0.992156742i$$

Steps Using the Quadratic Formula

Add 4y^{2} to both sides.

$$y+4+4y^{2}=0$$

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.

$$4y^{2}+y+4=0$$

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

$$y=\frac{-1±\sqrt{1^{2}-4\times 4\times 4}}{2\times 4}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-1±\sqrt{1-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$y=\frac{-1±\sqrt{1-64}}{2\times 4}$$

Add 1 to -64.

$$y=\frac{-1±\sqrt{-63}}{2\times 4}$$

Take the square root of -63.

$$y=\frac{-1±3\sqrt{7}i}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-1±3\sqrt{7}i}{8}$$

Now solve the equation y=\frac{-1±3\sqrt{7}i}{8} when ± is plus. Add -1 to 3i\sqrt{7}.

$$y=\frac{-1+3\sqrt{7}i}{8}$$

Now solve the equation y=\frac{-1±3\sqrt{7}i}{8} when ± is minus. Subtract 3i\sqrt{7} from -1.

$$y=\frac{-3\sqrt{7}i-1}{8}$$

The equation is now solved.

$$y=\frac{-1+3\sqrt{7}i}{8} y=\frac{-3\sqrt{7}i-1}{8}$$

Steps for Completing the Square

Add 4y^{2} to both sides.

$$y+4+4y^{2}=0$$

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

$$y+4y^{2}=-4$$

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.

$$4y^{2}+y=-4$$

Divide both sides by 4.

$$\frac{4y^{2}+y}{4}=-\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\frac{1}{4}y=-\frac{4}{4}$$

Divide -4 by 4.

$$y^{2}+\frac{1}{4}y=-1$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{1}{4}y+\left(\frac{1}{8}\right)^{2}=-1+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{1}{4}y+\frac{1}{64}=-1+\frac{1}{64}$$

Add -1 to \frac{1}{64}.

$$y^{2}+\frac{1}{4}y+\frac{1}{64}=-\frac{63}{64}$$

Factor y^{2}+\frac{1}{4}y+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{8}\right)^{2}=-\frac{63}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{63}{64}}$$

Simplify.

$$y+\frac{1}{8}=\frac{3\sqrt{7}i}{8} y+\frac{1}{8}=-\frac{3\sqrt{7}i}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$y=\frac{-1+3\sqrt{7}i}{8} y=\frac{-3\sqrt{7}i-1}{8}$$