Solve (2y)^2+y-3=0

$$( 2 y ) ^ { 2 } + y - 3 = 0$$

Solve for y

$$y=-1$$

$$y=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Expand \left(2y\right)^{2}.

$$2^{2}y^{2}+y-3=0$$

Calculate 2 to the power of 2 and get 4.

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

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.

$$a+b=1 ab=4\left(-3\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

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

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=-3 b=4$$

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

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

Factor out y in 4y^{2}-3y.

$$y\left(4y-3\right)+4y-3$$

Factor out common term 4y-3 by using distributive property.

$$\left(4y-3\right)\left(y+1\right)$$

To find equation solutions, solve 4y-3=0 and y+1=0.

$$y=\frac{3}{4} y=-1$$

Steps Using the Quadratic Formula

Expand \left(2y\right)^{2}.

$$2^{2}y^{2}+y-3=0$$

Calculate 2 to the power of 2 and get 4.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Square 1.

$$y=\frac{-1±\sqrt{1-4\times 4\left(-3\right)}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-1±\sqrt{1-16\left(-3\right)}}{2\times 4}$$

Multiply -16 times -3.

$$y=\frac{-1±\sqrt{1+48}}{2\times 4}$$

Add 1 to 48.

$$y=\frac{-1±\sqrt{49}}{2\times 4}$$

Take the square root of 49.

$$y=\frac{-1±7}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{-1±7}{8}$$

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

$$y=\frac{6}{8}$$

Reduce the fraction \frac{6}{8} to lowest terms by extracting and canceling out 2.

$$y=\frac{3}{4}$$

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

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

Divide -8 by 8.

$$y=-1$$

The equation is now solved.

$$y=\frac{3}{4} y=-1$$

Steps for Completing the Square

Expand \left(2y\right)^{2}.

$$2^{2}y^{2}+y-3=0$$

Calculate 2 to the power of 2 and get 4.

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

Add 3 to both sides. Anything plus zero gives itself.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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.

$$y^{2}+\frac{1}{4}y+\left(\frac{1}{8}\right)^{2}=\frac{3}{4}+\left(\frac{1}{8}\right)^{2}$$

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

$$y^{2}+\frac{1}{4}y+\frac{1}{64}=\frac{3}{4}+\frac{1}{64}$$

Add \frac{3}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{1}{4}y+\frac{1}{64}=\frac{49}{64}$$

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}.

$$\left(y+\frac{1}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

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

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

$$y=\frac{3}{4} y=-1$$