How do you solve $4{{x}^{2}}-6=74$?

Hint: In this question we will rearrange the terms in the expression by isolating the terms of $x$by sending the term $-6$ from the left-hand side to the right-hand side and then simplify the expression and take the square root of both the sides of the expression and simplify the root to get the required solution.

Complete step-by-step answer:
We have the expression given as:
$\Rightarrow 4{{x}^{2}}-6=74$
On transferring the term $-6$ from the left-hand side to the right-hand side, we get:
$\Rightarrow 4{{x}^{2}}=80$
On transferring the term $4$ from the left-hand side to the right-hand side, we get:
$\Rightarrow {{x}^{2}}=\dfrac{80}{4}$
On simplifying the fraction, we get:
$\Rightarrow {{x}^{2}}=20$
On taking the square root on both the sides, we get:
$\Rightarrow x=\pm \sqrt{20}$
Now we know that $20=4\times 5$ therefore, on substituting it in the expression, we get:
$\Rightarrow x=\pm \sqrt{4\times 5}$
Now we know that $4=2\times 2$ therefore, on substituting it in the expression, we get:
$\Rightarrow x=\pm \sqrt{2\times 2\times 5}$
Since the term $2$ is present two times in the in the square root, we can take it out from the square root as:
$\Rightarrow x=\pm 2\sqrt{5}$, which is the required solution.

Note: In this question we have the expression in the form of a polynomial equation with degree $2$. The degree of the expression is the highest value which is raised to an unknown variable in the equation, in this equation being $2$ which is on $x$.
It is to be remembered that when a term is transferred across the $=$ sign, if it is in addition or subtraction, it changes to subtraction and addition respectively. Same happens with terms which are in multiplication and subtraction.
It is also to be remembered that whenever the square root of a term is to be taken, two terms are written one which is positive and one which is negative.