2d^2-5d-4=0

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Solution for 2d^2-5d-4=0 equation: 



2d^2-5d-4=0

a = 2; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·2·(-4)
Δ = 57
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{-(-5)-\sqrt{57}}{2*2}=\frac{5-\sqrt{57}}{4}
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{57}}{2*2}=\frac{5+\sqrt{57}}{4}
$

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