Solve 4y^2=-8y-3

$$4 y ^ { 2 } = - 8 y - 3$$

Solve for y

$$y = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$y=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Add 8y to both sides.

$$4y^{2}+8y=-3$$

Add 3 to both sides.

$$4y^{2}+8y+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4y^{2}+ay+by+3. To find a and b, set up a system to be solved.

$$a+b=8 ab=4\times 3=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 8.

$$a=2 b=6$$

Rewrite 4y^{2}+8y+3 as \left(4y^{2}+2y\right)+\left(6y+3\right).

$$\left(4y^{2}+2y\right)+\left(6y+3\right)$$

Factor out 2y in the first and 3 in the second group.

$$2y\left(2y+1\right)+3\left(2y+1\right)$$

Factor out common term 2y+1 by using distributive property.

$$\left(2y+1\right)\left(2y+3\right)$$

To find equation solutions, solve 2y+1=0 and 2y+3=0.

$$y=-\frac{1}{2} y=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Add 8y to both sides.

$$4y^{2}+8y=-3$$

Add 3 to both sides.

$$4y^{2}+8y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\times 4\times 3}}{2\times 4}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\times 4\times 3}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-8±\sqrt{64-16\times 3}}{2\times 4}$$

Multiply -16 times 3.

$$y=\frac{-8±\sqrt{64-48}}{2\times 4}$$

Add 64 to -48.

$$y=\frac{-8±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$y=\frac{-8±4}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-8±4}{8}$$

Now solve the equation y=\frac{-8±4}{8} when ± is plus. Add -8 to 4.

$$y=-\frac{4}{8}$$

Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$y=-\frac{1}{2}$$

Now solve the equation y=\frac{-8±4}{8} when ± is minus. Subtract 4 from -8.

$$y=-\frac{12}{8}$$

Reduce the fraction \frac{-12}{8} to lowest terms by extracting and canceling out 4.

$$y=-\frac{3}{2}$$

The equation is now solved.

$$y=-\frac{1}{2} y=-\frac{3}{2}$$

Steps for Completing the Square

Add 8y to both sides.

$$4y^{2}+8y=-3$$

Divide both sides by 4.

$$\frac{4y^{2}+8y}{4}=-\frac{3}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\frac{8}{4}y=-\frac{3}{4}$$

Divide 8 by 4.

$$y^{2}+2y=-\frac{3}{4}$$

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

$$y^{2}+2y+1^{2}=-\frac{3}{4}+1^{2}$$

Square 1.

$$y^{2}+2y+1=-\frac{3}{4}+1$$

Add -\frac{3}{4} to 1.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

$$y=-\frac{1}{2} y=-\frac{3}{2}$$