(4v+3)(1+v)=0

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Solution for (4v+3)(1+v)=0 equation:



(4v+3)(1+v)=0
We add all the numbers together, and all the variables
(4v+3)(v+1)=0
We multiply parentheses ..
(+4v^2+4v+3v+3)=0
We get rid of parentheses
4v^2+4v+3v+3=0
We add all the numbers together, and all the variables
4v^2+7v+3=0
a = 4; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*4}=\frac{-8}{8} =-1 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*4}=\frac{-6}{8} =-3/4 $

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