(2v+3)(6-v)=0

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Solution for (2v+3)(6-v)=0 equation:



(2v+3)(6-v)=0
We add all the numbers together, and all the variables
(2v+3)(-1v+6)=0
We multiply parentheses ..
(-2v^2+12v-3v+18)=0
We get rid of parentheses
-2v^2+12v-3v+18=0
We add all the numbers together, and all the variables
-2v^2+9v+18=0
a = -2; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·(-2)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*-2}=\frac{-24}{-4} =+6 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*-2}=\frac{6}{-4} =-1+1/2 $

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