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9=r2+3
We move all terms to the left:
9-(r2+3)=0
We add all the numbers together, and all the variables
-(+r^2+3)+9=0
We get rid of parentheses
-r^2-3+9=0
We add all the numbers together, and all the variables
-1r^2+6=0
a = -1; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·(-1)·6
Δ = 24
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-1}=\frac{0-2\sqrt{6}}{-2} =-\frac{2\sqrt{6}}{-2} =-\frac{\sqrt{6}}{-1} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-1}=\frac{0+2\sqrt{6}}{-2} =\frac{2\sqrt{6}}{-2} =\frac{\sqrt{6}}{-1} $
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