How to solve the equation $(2x + 1)(2x + 3) = 143$ without using the Quadratic Formula? I have been a bit stuck on this question.
The product of two consecutive odd numbers is $143$. Find the next numbers.
I have made this into:
$$
(2x+1)(2x+3)=143.
$$
I got $x_1 = -7$ and $x_2 = -5$ for this, which doesn't seem right.
How can I fix this? (By the way we are only able to solve for $x$ by using factorising and not with the quadratic formula).
 A: An idea: decompose in primes product $\;143=11\cdot13\;$ . 
Observe that we're given $\;(2x+1)(2x+3)=143\;$ , and the left side is the product of two consecutive odd integers. Once you get the prime product for $\;143\;$ you already have not many choices: one must be $\;11\;$ , and the other one $\;13\;$ ...or both negative: $\;-11,\,-13\;$ , and you get:
$$\begin{cases}2x+1=11\\2x+3=13\end{cases}\implies \color{red}{x=5}\;,\;\;\begin{cases}2x+1=-13\\2x+3=-11\end{cases}\implies \color{red}{x=-7}$$
A: \begin{align*}
(2x + 1)(2x + 3) & = 143\\
2x(2x + 3) + 1(2x + 3) & = 143 && \text{expand}\\
4x^2 + 6x + 2x + 3 & = 143 && \text{apply the distributive law}\\
4x^2 + 8x + 3 & = 143 && \text{combine like terms}\\
4x^2 + 8x - 140 & = 0 && \text{subtract $140$ from each side of the equation}\\
x^2 + 2x - 35 & = 0 && \text{divide each side of the equation by $4$}\\
(x + 7)(x - 5) & = 0 && \text{factor the quadratic}
\end{align*}
If a product is equal to zero, then one of the factors must be equal to zero.  Hence, 
\begin{align*}
x + 7 & = 0 & x - 5 & = 0\\
x & = -7 & x & = 5
\end{align*}
You can verify that these values are correct by substituting them into the equation $(2x + 1)(2x + 3) = 143$.
If $x = -7$, then the consecutive odd numbers are $2(-7) + 1 = -13$ and $2(-7) + 3 = -11$.  
If $x = 5$, then the consecutive odd numbers are $2(5) + 1 = 11$ and $2(5) + 3 = 13$. 
A: $$(2x + 1)(2x + 3) = 143$$
Let $y=\frac{(2x+1)+(2x+3)}{2}=2x+2$. Then:
$$(y-1)(y+1)=143\implies y^2-1=143=144-1$$
Thus, $y^2=144\implies y=\pm12\implies(2x+2)=\pm12\implies x\in\{-7,5\}$
