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5d^2+19d+16=0
a = 5; b = 19; c = +16;
Δ = b2-4ac
Δ = 192-4·5·16
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{41}}{2*5}=\frac{-19-\sqrt{41}}{10} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{41}}{2*5}=\frac{-19+\sqrt{41}}{10} $
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