(4y+2)²-(y+1)(y-1)=5y(2+3y)+y
You asked:
Solve the equation \({\left( 4 y + 2 \right)}^{2} - \left(y + 1\right) \left(y - 1\right) = 5 \cdot y\left( 2 + 3 y \right) + y\) for the variable \(y\).
MathBot Answer:
The solution is: \[y = -1\]
\[\begin{aligned}15 y^{2} + 16 y + 5 &= 15 y^{2} + 11 y\\5 y &= -5\\y &= -1\end{aligned}\]