How to solve $\frac{99.5−0.8n}{\sqrt{0.16n}}=-1.645$?

Question

$\frac{99.5−0.8n}{\sqrt{0.16n}}=-1.645$

Simplifying and I get

$99.5−0.8n+1.645\sqrt{0.16} n^{1/2}=0$

$0.8n-1.645\sqrt{0.16} n^{1/2}-99.5=0$

Can't use quadratic formula.

Answer 1

The idea you had was on the right track, you just need to continue by squaring both sides. $$\frac{99.5-0.8n}{\sqrt {0.16n}}=-1.645$$ $$99.5-0.8n=-1.645 \sqrt {0.16n}$$ $$(99.5-0.8n)^2 = (-1.645 \sqrt {0.16n})^2$$ $$9900.25-159.2n+0.64n^2=.432964n$$ $$9900.25-159.632964n+0.64n^2=0$$ From this, you can use any methods you know, such as the quadratic formula.

Answer 2

Instead of the absurdly specific $\dfrac{99.5−0.8n}{\sqrt{0.16n}}=-1.645 $, consider $\frac{a−bn}{\sqrt{cn}}=-d $ with $a, b, c, d > 0$.

Then $a-bn =-d\sqrt{n} $. Squaring, and remembering that we have to check the roots because of this, $a^2-2abn+b^2n^2 =d^2n $ or $b^2n^2-(2ab+d^2)n+a^2 = 0$.

Applying the standard quadratic formula,

$\begin{array}\\ n &=\dfrac{(2ab+d^2)\pm \sqrt{(2ab+d^2)^2-4a^2b^2}}{2b^2}\\ &=\dfrac{(2ab+d^2)\pm \sqrt{4a^2b^2+4abd^2+d^4-4a^2b^2}}{2b^2}\\ &=\dfrac{(2ab+d^2)\pm \sqrt{4abd^2+d^4}}{2b^2}\\ &=\dfrac{(2ab+d^2)\pm d\sqrt{4ab+d^2}}{2b^2}\\ \end{array} $

Since $4ab+d^2 > 0$, the roots are real; since $(2ab+d^2)^2 \gt 4abd^2+d^4 $, both roots are positive.

Now substitute your parameters.