How do you solve \[10{v^2} = 13v + 3\]?

Hint: After rearranging the given equation we will have a polynomial equation. The obtained polynomial is of degree 2. Instead of ‘x’ as a variable we have ‘v’ as a variable. We can solve this by using factorization methods or by using quadratic formulas. We use quadratic formula if factorization fails. We know that a polynomial equation has exactly as many roots as its degree.

Complete step by step answer:
Given,
\[10{v^2} = 13v + 3\]
Rearranging we have,
\[10{v^2} - 13v - 3 = 0\]
On comparing the given equation with the standard quadratic equation \[A{v^2} + Bv + C = 0\]. Where ‘A’ and ‘B’ are coefficients of \[{v^2}\] and coefficient of ‘v’ respectively.
We have \[A = 10\], \[B = - 13\] and \[C = - 3\].
The standard form of the factorization of quadratic equation is \[A{v^2} + {B_1}v + {B_2}v + C = 0\], which satisfies the condition \[{B_1} \times {B_2} = A \times C\] and \[{B_1} + {B_2} = B\].
We can write the given equation as \[10{v^2} - 15v + 2v - 3 = 0\], where \[{B_1} = - 15\] and \[{B_2} = 2\]. Also \[{B_1} \times {B_2} = ( - 15) \times (2) = - 30(A \times C)\] and \[{B_1} + {B_2} = ( - 15) + (2) = - 13(B)\].
\[10{v^2} - 13v - 3 \Rightarrow 10{v^2} - 15v + 2v - 3 = 0\]
\[10{v^2} - 15v + 2v - 3 = 0\]
In the first two terms we take ‘5v’ as common and in the remaining term we take 1 as common,
\[5v(2v - 3) + 1(2v - 3) = 0\]
Again taking \[(2v - 3)\] as common we have,
\[(2v - 3)(5v + 1) = 0\]
By zero multiplication property we have,
\[(2v - 3) = 0\] and \[(5v + 1) = 0\].
\[2v = 3\] and \[5v = - 1\]
\[ \Rightarrow v = \dfrac{3}{2}\] and \[v = - \dfrac{1}{5}\].
These are the roots of the given problem.

Note: In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts.