74w^2+48w=0

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Solution for 74w^2+48w=0 equation:



74w^2+48w=0
a = 74; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·74·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*74}=\frac{-96}{148} =-24/37 $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*74}=\frac{0}{148} =0 $

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