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(2/3)r=11
We move all terms to the left:
(2/3)r-(11)=0
Domain of the equation: 3)r!=0We add all the numbers together, and all the variables
r!=0/1
r!=0
r∈R
(+2/3)r-11=0
We multiply parentheses
2r^2-11=0
a = 2; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·2·(-11)
Δ = 88
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{22}}{2*2}=\frac{0-2\sqrt{22}}{4} =-\frac{2\sqrt{22}}{4} =-\frac{\sqrt{22}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{22}}{2*2}=\frac{0+2\sqrt{22}}{4} =\frac{2\sqrt{22}}{4} =\frac{\sqrt{22}}{2} $
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