(7-v)(5v+3)=0

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Solution for (7-v)(5v+3)=0 equation:



(7-v)(5v+3)=0
We add all the numbers together, and all the variables
(-1v+7)(5v+3)=0
We multiply parentheses ..
(-5v^2-3v+35v+21)=0
We get rid of parentheses
-5v^2-3v+35v+21=0
We add all the numbers together, and all the variables
-5v^2+32v+21=0
a = -5; b = 32; c = +21;
Δ = b2-4ac
Δ = 322-4·(-5)·21
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-38}{2*-5}=\frac{-70}{-10} =+7 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+38}{2*-5}=\frac{6}{-10} =-3/5 $

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