Answer:
the solutions are 5 and -5
Step-by-step explanation:
[tex]5a^2-44=81[/tex]
first leave it as
[tex]something = 0[/tex]
so working on
[tex]5a^2-44-81=81-81\\\\5a^2-44-81=0[/tex]
add like terms
[tex]5a^2-125=0[/tex]
get common factor (in this case 5)
[tex]5(a^2-25)=0[/tex]
divide by 5
[tex]\frac{1}{5}[5(a^2-25)]=\frac{1}{5} [0][/tex]
simplify again
[tex]1(a^2-25)=0[/tex]
since 1*x = x we have
[tex]a^2-25=0[/tex]
now use the difference of squares formula [tex]x^2-y^2=(x-y)(x+y)[/tex]
*note that 25 = 5^2
[tex](a-5) (a+5)=0[/tex]
now we have two cases
[tex]a-5=0[/tex]
and
[tex]a+5=0[/tex]
solve for each one
[tex]a-5=0\\\\a-5+5=0+5\\\\a+0=5\\\\a=5[/tex]
and the other one is
[tex]a+5=0\\\\a+5-5=0-5\\\\a+0=-5\\\\a=-5[/tex]
so the solutions are 5 and -5
