(2v^2+7v+3)=0

Solution for (2v^2+7v+3)=0 equation:

(2v^2+7v+3)=0

We get rid of parentheses

2v^2+7v+3=0

a = 2; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·2·3
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-5}{2*2}=\frac{-12}{4}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+5}{2*2}=\frac{-2}{4}
=-1/2
$