r^2=3/16

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Solution for r^2=3/16 equation:



r^2=3/16
We move all terms to the left:
r^2-(3/16)=0
We add all the numbers together, and all the variables
r^2-(+3/16)=0
We get rid of parentheses
r^2-3/16=0
We multiply all the terms by the denominator
r^2*16-3=0
Wy multiply elements
16r^2-3=0
a = 16; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·16·(-3)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*16}=\frac{0-8\sqrt{3}}{32} =-\frac{8\sqrt{3}}{32} =-\frac{\sqrt{3}}{4} $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*16}=\frac{0+8\sqrt{3}}{32} =\frac{8\sqrt{3}}{32} =\frac{\sqrt{3}}{4} $

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