How do you factor \[2{v^2} + 11v + 5\] ?

Hint: Here we are given with a quadratic equation. To solve the quadratic equation is nothing but finding the factors of the equation. Using a quadratic formula is finding the roots with the help of discriminant. We will compare the given quadratic equation with the general quadratic equation of the form \[a{x^2} + bx + c = 0\] . So let’s start solving!
Quadratic formula: \[ \Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

Complete step-by-step answer:
Given that \[2{v^2} + 11v + 5\] is the equation given.
Equate this to zero first \[2{v^2} + 11v + 5 = 0\]
Now comparing it with the general equation \[a{x^2} + bx + c = 0\] we get \[a = 2,b = 11\& c = 5\] .
Putting these values in quadratic formula,
 \[ \Rightarrow \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
 \[ \Rightarrow \dfrac{{ - \left( {11} \right) \pm \sqrt {{{\left( {11} \right)}^2} - 4 \times 2 \times 5} }}{{2 \times 2}}\]
Now on solving the brackets,
 \[ \Rightarrow \dfrac{{ - 11 \pm \sqrt {121 - 40} }}{4}\]
 \[ \Rightarrow \dfrac{{ - 11 \pm \sqrt {81} }}{4}\]
We know that 81 is the perfect square of 9. Thus taking the roots we get,
 \[ \Rightarrow \dfrac{{ - 11 \pm 9}}{4}\]
Separating the roots we get,
 \[ \Rightarrow \dfrac{{ - 11 - 9}}{4}or \Rightarrow \dfrac{{ - 11 + 9}}{4}\]
On solving this we get,
 \[ \Rightarrow \dfrac{{ - 20}}{4}or \Rightarrow \dfrac{{ - 2}}{4}\]
Dividing this,
 \[ \Rightarrow - 5or \Rightarrow \dfrac{{ - 1}}{2}\]
This is our answer or we can say the roots or factors are \[ \Rightarrow - 5\; or\; \Rightarrow \dfrac{{ - 1}}{2}\]
So, the correct answer is “ \[ - 5 \;or \; \dfrac{{ - 1}}{2}\] ”.

Note: Note that quadratic formula is used to find the roots of a given quadratic equation. Sometimes we can factorize the roots directly. But quadratic formulas can be used generally to find the roots of any quadratic equation. The value of discriminant is used to decide the type of roots so obtained such that roots are equal or different and are real or not.