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2d^2+11d-420=0
a = 2; b = 11; c = -420;
Δ = b2-4ac
Δ = 112-4·2·(-420)
Δ = 3481
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{3481}=59$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-59}{2*2}=\frac{-70}{4} =-17+1/2 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+59}{2*2}=\frac{48}{4} =12 $
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