Question

solve the equation using the quadratic formula. express your answer in exact, simplified form.
$y^{2}=2y + 4$
the solution set is {}

Explanation:

Step1: Rewrite in standard form

$y^{2}-2y - 4=0$. Here $a = 1$, $b=-2$, $c = - 4$.

Step2: Apply quadratic formula

The quadratic formula is $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b=-2$, $c = - 4$ into it: $y=\frac{-(-2)\pm\sqrt{(-2)^{2}-4\times1\times(-4)}}{2\times1}$.

Step3: Simplify the expression

First, calculate inside the square - root: $(-2)^{2}-4\times1\times(-4)=4 + 16=20$. Then, $\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$. The expression becomes $y=\frac{2\pm2\sqrt{5}}{2}$. Factor out a 2 in the numerator: $y=\frac{2(1\pm\sqrt{5})}{2}=1\pm\sqrt{5}$.

Answer:

$\{1 + \sqrt{5},1-\sqrt{5}\}$