+0

+1
32
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+966

Solve for g

h=$$\sqrt{\frac{g^2}{d}-5t}-1$$

ManuelBautista2019  Mar 4, 2018

#1
+6608
+1

$$h=\sqrt{\frac{g^2}{d}-5t}-1$$

Add  1  to both sides of the equation.

$$h+1=\sqrt{\frac{g^2}{d}-5t}$$

Square both sides of the equation.

$$(h+1)^2=\frac{g^2}{d}-5t$$

$$(h+1)^2+5t=\frac{g^2}{d}$$

Multiply both sides by  d .

$$d[(h+1)^2+5t]=g^2$$

Take the ± square root of both sides.

$$±\sqrt{d[(h+1)^2+5t]}=g \\~\\ g=±\sqrt{d[(h+1)^2+5t]}$$

hectictar  Mar 4, 2018
Sort:

#1
+6608
+1

$$h=\sqrt{\frac{g^2}{d}-5t}-1$$

Add  1  to both sides of the equation.

$$h+1=\sqrt{\frac{g^2}{d}-5t}$$

Square both sides of the equation.

$$(h+1)^2=\frac{g^2}{d}-5t$$

$$(h+1)^2+5t=\frac{g^2}{d}$$

Multiply both sides by  d .

$$d[(h+1)^2+5t]=g^2$$

Take the ± square root of both sides.

$$±\sqrt{d[(h+1)^2+5t]}=g \\~\\ g=±\sqrt{d[(h+1)^2+5t]}$$

hectictar  Mar 4, 2018
#2
+966
+2

Thank you!!

ManuelBautista2019  Mar 4, 2018

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