10v^2+41v+21=0

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Solution for 10v^2+41v+21=0 equation:



10v^2+41v+21=0
a = 10; b = 41; c = +21;
Δ = b2-4ac
Δ = 412-4·10·21
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-29}{2*10}=\frac{-70}{20} =-3+1/2 $
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+29}{2*10}=\frac{-12}{20} =-3/5 $

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