d^2+38d=0

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Solution for d^2+38d=0 equation: 



d^2+38d=0

a = 1; b = 38; c = 0;
Δ = b2-4ac
Δ = 382-4·1·0
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$


$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-38}{2*1}=\frac{-76}{2}
=-38
$


$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+38}{2*1}=\frac{0}{2}
=0
$

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