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20d^2-3d-2=0
a = 20; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·20·(-2)
Δ = 169
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*20}=\frac{-10}{40} =-1/4 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*20}=\frac{16}{40} =2/5 $
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