23v^2+47v=0

Solution for 23v^2+47v=0 equation:

23v^2+47v=0

a = 23; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·23·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*23}=\frac{-94}{46}
=-2+1/23
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*23}=\frac{0}{46}
=0
$