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r^2-2r-63=0
a = 1; b = -2; c = -63;
Δ = b2-4ac
Δ = -22-4·1·(-63)
Δ = 256
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{256}=16$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*1}=\frac{-14}{2} =-7 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*1}=\frac{18}{2} =9 $
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