Question
$$(n+1)^2-(n+2)^2 = 0$$
Answer
$$ n = -\dfrac{3}{2} $$
Explanation
$$ \begin{aligned} (n+1)^2-(n+2)^2 &= 0&& \text{simplify left side} \\[1 em]n^2+2n+1-(1n^2+4n+4) &= 0&& \\[1 em]n^2+2n+1-n^2-4n-4 &= 0&& \\[1 em]n^2+2n+1-n^2-4n-4 &= 0&& \\[1 em]-2n-3 &= 0&& \text{ move the constants to the right } \\[1 em]-2n &= 3&& \text{ divide both sides by $ -2 $ } \\[1 em]n &= -\frac{3}{2}&& \\[1 em] \end{aligned} $$
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