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11r^2+7r-3=0
a = 11; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·11·(-3)
Δ = 181
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{-(7)-\sqrt{181}}{2*11}=\frac{-7-\sqrt{181}}{22} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{181}}{2*11}=\frac{-7+\sqrt{181}}{22} $
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