5v^2+29v+29=0

Solution for 5v^2+29v+29=0 equation:

5v^2+29v+29=0

a = 5; b = 29; c = +29;
Δ = b2-4ac
Δ = 292-4·5·29
Δ = 261
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{261}=\sqrt{9*29}=\sqrt{9}*\sqrt{29}=3\sqrt{29}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-3\sqrt{29}}{2*5}=\frac{-29-3\sqrt{29}}{10}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+3\sqrt{29}}{2*5}=\frac{-29+3\sqrt{29}}{10}
$