How do I solve ot √ (20 + √ (20 + √ (20 + √ (20...)))) =?

 How do I solve ot √ (20 + √ (20 + √ (20 + √ (20...)))) =?💝👇

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To solve the equation involving an infinite nested radical:

$$𝑥=\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+\cdots }}}}$$

we start by recognizing that the expression inside the square root is the same as the whole expression itself. This gives us the equation:

$$𝑥=\sqrt{20+𝑥}$$

Next, we square both sides of the equation to eliminate the square root:

$${𝑥}^2=20+𝑥$$

Now, we rearrange the equation to form a standard quadratic equation:

$${𝑥}^2-𝑥-20=0$$

We solve this quadratic equation using the quadratic formula $𝑥=\frac{-𝑏\pm \sqrt{{𝑏}^2-4𝑎𝑐}}{2𝑎}$. Here, $𝑎=1$, $𝑏=-1$, and $𝑐=-20$. Plugging these values into the formula, we get:

$$𝑥=\frac{-(-1)\pm \sqrt{(-1{)}^2-4\cdot 1\cdot (-20)}}{2\cdot 1}$$ $$𝑥=\frac{1\pm \sqrt{1+80}}{2}$$ $$𝑥=\frac{1\pm \sqrt{81}}{2}$$ $$𝑥=\frac{1\pm 9}{2}$$

This gives us two solutions:

$$𝑥=\frac{1+9}{2}=\frac{10}{2}=5$$ $$𝑥=\frac{1-9}{2}=\frac{-8}{2}=-4$$