v^2-14v+40=0

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Solution for v^2-14v+40=0 equation: 



v^2-14v+40=0

a = 1; b = -14; c = +40;
Δ = b2-4ac
Δ = -142-4·1·40
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*1}=\frac{8}{2}
=4
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*1}=\frac{20}{2}
=10
$

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