(3x+4)(2x-1)+4=0

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Solution for (3x+4)(2x-1)+4=0 equation:



(3x+4)(2x-1)+4=0
We multiply parentheses ..
(+6x^2-3x+8x-4)+4=0
We get rid of parentheses
6x^2-3x+8x-4+4=0
We add all the numbers together, and all the variables
6x^2+5x=0
a = 6; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·6·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*6}=\frac{-10}{12} =-5/6 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*6}=\frac{0}{12} =0 $

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