v^2-5+3v=0

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Solution for v^2-5+3v=0 equation: 



v^2-5+3v=0

a = 1; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·1·(-5)
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{29}}{2*1}=\frac{-3-\sqrt{29}}{2}
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*1}=\frac{-3+\sqrt{29}}{2}
$

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