Use Square Roots to Solve Quadratic Equations
Key Questions

If the question is about using the square root directly against the equation, the answer is definitely NO.
However, with certain transformation of a given equation into a different but equivalent form it is possible. Here is the idea.
Assume, for example, the same equation as analyzed in the previous answer:
#x^2+x=63# If we could transform it to something like
#y^2=b# then the square root of both sides would deliver a solution.
So, let's transform our equation to this form.
Expression#x^2+x# is not a square of anything, but#x^2+x+1/4# is a square of#x+1/2# because
#(x+1/2)^2=x^2+2*x*1/2+1/4=x^2+x+1/4# Therefore, it is reasonable to transform the original equation into
#(x+1/2)^21/4=63# or
#(x+1/2)^2=253/4#
From the last equation, which is absolutely equivalent to the original one, using the operation of the square root, we derive two linear equations:
#x+1/2=sqrt(253)/2# and#x+1/2=sqrt(253)/2# So, two solutions are:
#x=(1+sqrt(253))/2# and#x=(1sqrt(253))/2# The above method is pretty universal and handy if you don't remember a formula for solutions of a quadratic equation. Let me illustrate this with another example.
#3x^2+2x+8=0# Step 1. Divide everything by
#3# to have#x^2# with a multiplier#1# :
#x^22/3x8/3=0# Step 2. Since a coefficient at
#x# is#2/3# , use#(x1/3)^2# in a transformed equation:
#(x1/3)^21/98/3=0# or
#(x1/3)^2=25/9# Step 3. Use square root:
#x1/3=5/3# and#x1/3=5/3# Step 4. Solutions:
#x=6/3=2# and#x=4/3# 
Let us solve the following quadratic equation.
#x^26x+7=0# by adding 2,
#x^26x+9=2# #(x3)^2=2# by taking the squareroot,
#x3=pm sqrt{2}# by adding 3,
#x=3 pm sqrt{2}#
I hope that this was helpful.
Questions
Quadratic Equations and Functions

Quadratic Functions and Their Graphs

Vertical Shifts of Quadratic Functions

Use Graphs to Solve Quadratic Equations

Use Square Roots to Solve Quadratic Equations

Completing the Square

Vertex Form of a Quadratic Equation

Quadratic Formula

Comparing Methods for Solving Quadratics

Solutions Using the Discriminant

Linear, Exponential, and Quadratic Models

Applications of Function Models