(2v+5)(1-v)=0

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Solution for (2v+5)(1-v)=0 equation:



(2v+5)(1-v)=0
We add all the numbers together, and all the variables
(2v+5)(-1v+1)=0
We multiply parentheses ..
(-2v^2+2v-5v+5)=0
We get rid of parentheses
-2v^2+2v-5v+5=0
We add all the numbers together, and all the variables
-2v^2-3v+5=0
a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4} =1 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4} =-2+1/2 $

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