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r^2-45r+168=0
a = 1; b = -45; c = +168;
Δ = b2-4ac
Δ = -452-4·1·168
Δ = 1353
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{-(-45)-\sqrt{1353}}{2*1}=\frac{45-\sqrt{1353}}{2} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{1353}}{2*1}=\frac{45+\sqrt{1353}}{2} $
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