(3x+4)(2x-0)=0

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Solution for (3x+4)(2x-0)=0 equation:



(3x+4)(2x-0)=0
We add all the numbers together, and all the variables
(3x+4)(+2x)=0
We multiply parentheses ..
(+6x^2+8x)=0
We get rid of parentheses
6x^2+8x=0
a = 6; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·6·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*6}=\frac{-16}{12} =-1+1/3 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*6}=\frac{0}{12} =0 $

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