16y^2+21y=0

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Solution for 16y^2+21y=0 equation:



16y^2+21y=0
a = 16; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·16·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*16}=\frac{-42}{32} =-1+5/16 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*16}=\frac{0}{32} =0 $

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