21v^2-34v+8=0

Solution for 21v^2-34v+8=0 equation:

21v^2-34v+8=0

a = 21; b = -34; c = +8;
Δ = b2-4ac
Δ = -342-4·21·8
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-22}{2*21}=\frac{12}{42}
=2/7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+22}{2*21}=\frac{56}{42}
=1+1/3
$