How do you solve \[5{q^2} + 18q = 8?\]

Hint: In the given question we have been asked to find the solutions of a quadratic equation \[5{q^2} + 18q = 8\]. Now as we know that the standard form of a quadratic equation is \[a{x^2} + bx + c = 0\]. So first convert the given term in standard form and then solve the quadratic equation by using factorization formula to find the solution of the quadratic equation.

Complete step by step solution:
First of all convert the given term into basic quadratic equation for which is \[a{x^2} + bx + c = 0\]
We have \[5{q^2} + 18q = 8\]
Now subtracting 8 from both the sides LHS as well as from RHS we get:
\[
  5{q^2} + 18q - 8 = 8 - 8 \\
   \Rightarrow 5{q^2} + 18q - 8 = 0 \\
 \]
Now multiplying the c and a terms of this quadratic equation we get \[ - 40\]
Now see what factors of\[ - 40\] will add to \[18\].
So by hit and trial we get that \[ - 2\] and \[20\] are the factors that add to \[18\]
So we can expand our quadratic equation to:
\[5{q^2} + 20q - 2q - 8 = 0\]
Now find the GCF of first two and last two terms and take the GCF common from both i.e.
\[5q(q + 4) - 2(q + 4) = 0\]
Now you can see that \[(q - 4)\] is common in both the terms so we can further write our equation as:
\[(q - 4)(5q - 2) = 0\]
Now since we need to find the value of \[q\] for which the above term is equal to \[0\] so we will put both the terms equal to \[0\] i.e.
\[
  (q - 4) = 0 \\
   \Rightarrow (5q - 2) = 0 \\
 \]
 Further solving for \[q\] we get: \[q = 4,q = \dfrac{2}{5}\]

Hence, the roots of the quadratic equation \[5{q^2} + 18q = 8\] are \[4\] and \[\dfrac{2}{5}\].

Note: There are many methods to solve a quadratic equation but factorization method which we used in the above question is fast and easy method to solve quadratic equation but you should remember that if the quadratic equation have irrational solution then you should not use the above given method, basically this method is used only when quadratic equation have some rational numbers as its solution.