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r^2+15r-195=0
a = 1; b = 15; c = -195;
Δ = b2-4ac
Δ = 152-4·1·(-195)
Δ = 1005
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}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{1005}}{2*1}=\frac{-15-\sqrt{1005}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{1005}}{2*1}=\frac{-15+\sqrt{1005}}{2} $
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