Question

solve for y.
$2y^2 - 3y - 5 = (y + 1)^2$
if there is more than one solution, separate them with commas.
$y = \boxed{}$

Explanation:

Step1: Expand the right - hand side

We know that \((a + b)^2=a^{2}+2ab + b^{2}\). For \((y + 1)^2\), we have \(a = y\) and \(b = 1\), so \((y + 1)^2=y^{2}+2y + 1\).
The original equation \(2y^{2}-3y - 5=(y + 1)^2\) becomes \(2y^{2}-3y - 5=y^{2}+2y + 1\).

Step2: Move all terms to the left - hand side

Subtract \(y^{2}\), \(2y\) and \(1\) from both sides of the equation:
\(2y^{2}-3y - 5-(y^{2}+2y + 1)=0\)
Simplify the left - hand side:
\(2y^{2}-3y - 5 - y^{2}-2y - 1 = 0\)
Combine like terms: \((2y^{2}-y^{2})+(-3y-2y)+(-5 - 1)=0\)
Which gives \(y^{2}-5y - 6 = 0\).

Step3: Factor the quadratic equation

We need to find two numbers that multiply to \(-6\) and add up to \(-5\). The numbers are \(-6\) and \(1\) since \((-6)\times1=-6\) and \(-6 + 1=-5\).
So, \(y^{2}-5y - 6=(y - 6)(y+1)=0\).

Step4: Solve for y

Set each factor equal to zero:

• If \(y - 6=0\), then \(y = 6\).
• If \(y + 1=0\), then \(y=-1\).

Answer:

\(y=-1,6\)