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r2=600
We move all terms to the left:
r2-(600)=0
We add all the numbers together, and all the variables
r^2-600=0
a = 1; b = 0; c = -600;
Δ = b2-4ac
Δ = 02-4·1·(-600)
Δ = 2400
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*1}=\frac{0-20\sqrt{6}}{2} =-\frac{20\sqrt{6}}{2} =-10\sqrt{6} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*1}=\frac{0+20\sqrt{6}}{2} =\frac{20\sqrt{6}}{2} =10\sqrt{6} $
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