How do you solve #\frac { 1} { 4} x ^ { 2} + 4x - 3= 0#?

Use delta formula:
The delta formula says:
# ax^2+bx+c=0 => x= \frac{-b+-\sqrt(\Delta)}{2a}#
where #\Delta = b^2-4ac#

and just put the values in the formula:
#\Delta = 4^2-4\frac{1}{4}\times-3 = 16 + 3 = 19#
and #x# will be:

#x = \frac{-4+-\sqrt(19)}{2\times\frac{1}{4}} = 2(-4+-\sqrt(19))#

There are two answers:
#x_1 = 2(-4+\sqrt(19)) #
#x_2 = 2(-4-\sqrt(19)) #

Given
#color(white)("XXX")1/4x^2+4x-3=0#

This becomes easier if we multiply both sides by #4# to get rid of the fraction:
#color(white)("XXX")x^2+16x-12=0#

We can then solve this using the quadratic equation or by completing the square method.

Completing the square

#x^2+16x-12=0#

#color(white)("XXX")rarr x^2+16x=12#

Noting that if #x^2+16x# are the first 2 terms of a squared binomial, then the third term must be #(16/2)^2=8^2 (=64)#
#color(white)("XXX")rarr x^2+16x+8^2=12+64#

#color(white)("XXX")rarr (x+8)^2=76#

#color(white)("XXX")rarr x+8=+-sqrt(76)=+-2sqrt(19)#

#color(white)("XXX")rarr x=-8+-2sqrt(19)#

Quadratic formula

#ax^2+bx+c=0 rarr x=(-b+-sqrt(b^2+4ac))/(2a)#

In this case
#1x^2+16x-12=0 rarr x=(-16+-sqrt(16^2-4 * 1 * (-12)))/(2 * 1)#

#color(white)("XXXXXXXXXXXXXXX")=(-16+sqrt(304))/2#

#color(white)("XXXXXXXXXXXXXXX")=(-16+4sqrt(19))/2#

#color(white)("XXXXXXXXXXXXXXX")=-8+-2sqrt(19)#