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2r^2+13r+16=0
a = 2; b = 13; c = +16;
Δ = b2-4ac
Δ = 132-4·2·16
Δ = 41
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{-(13)-\sqrt{41}}{2*2}=\frac{-13-\sqrt{41}}{4} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{41}}{2*2}=\frac{-13+\sqrt{41}}{4} $
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