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r2-2r=35
We move all terms to the left:
r2-2r-(35)=0
We add all the numbers together, and all the variables
r^2-2r-35=0
a = 1; b = -2; c = -35;
Δ = b2-4ac
Δ = -22-4·1·(-35)
Δ = 144
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*1}=\frac{-10}{2} =-5 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*1}=\frac{14}{2} =7 $
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