h=-5(k-3)^2+47 -5(t-3)^2+47=20

## Step 1: Isolate the squared term<br />### Subtract 47 from both sides of the equation $-5(t-3)^2 + 47 = 20$.<br />$-5(t-3)^2 = 20 - 47$<br />$-5(t-3)^2 = -27$<br /><br />## Step 2: Divide by the coefficient of the squared term<br />### Divide both sides by -5.<br />$(t-3)^2 = \frac{-27}{-5}$<br />$(t-3)^2 = \frac{27}{5}$<br /><br />## Step 3: Take the square root of both sides<br />### Take the square root of both sides. Remember to consider both positive and negative roots.<br />$t-3 = \pm\sqrt{\frac{27}{5}}$<br /><br />## Step 4: Simplify the square root<br />### Simplify the square root.<br />$t-3 = \pm\frac{\sqrt{27}}{\sqrt{5}} = \pm\frac{3\sqrt{3}}{\sqrt{5}}$<br />$t-3 = \pm\frac{3\sqrt{3}\sqrt{5}}{\sqrt{5}\sqrt{5}} = \pm\frac{3\sqrt{15}}{5}$<br /><br />## Step 5: Isolate t<br />### Add 3 to both sides.<br />$t = 3 \pm \frac{3\sqrt{15}}{5}$