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r^2+2r=9
We move all terms to the left:
r^2+2r-(9)=0
a = 1; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·1·(-9)
Δ = 40
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*1}=\frac{-2-2\sqrt{10}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*1}=\frac{-2+2\sqrt{10}}{2} $
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