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12=r^2+2r
We move all terms to the left:
12-(r^2+2r)=0
We get rid of parentheses
-r^2-2r+12=0
We add all the numbers together, and all the variables
-1r^2-2r+12=0
a = -1; b = -2; c = +12;
Δ = b2-4ac
Δ = -22-4·(-1)·12
Δ = 52
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*-1}=\frac{2-2\sqrt{13}}{-2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*-1}=\frac{2+2\sqrt{13}}{-2} $
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