(d+5)(4d+1)=0

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Solution for (d+5)(4d+1)=0 equation:



(d+5)(4d+1)=0
We multiply parentheses ..
(+4d^2+d+20d+5)=0
We get rid of parentheses
4d^2+d+20d+5=0
We add all the numbers together, and all the variables
4d^2+21d+5=0
a = 4; b = 21; c = +5;
Δ = b2-4ac
Δ = 212-4·4·5
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*4}=\frac{-40}{8} =-5 $
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*4}=\frac{-2}{8} =-1/4 $

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