22x^2+48x=0

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Solution for 22x^2+48x=0 equation:



22x^2+48x=0
a = 22; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·22·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*22}=\frac{-96}{44} =-2+2/11 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*22}=\frac{0}{44} =0 $

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