23v^2+47v+2=0

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

23v^2+47v+2=0

a = 23; b = 47; c = +2;
Δ = b2-4ac
Δ = 472-4·23·2
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-45}{2*23}=\frac{-92}{46}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+45}{2*23}=\frac{-2}{46}
=-1/23
$