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Verified Answer :

deannas4395

First, you have to get the equation equal to 0. To do that, you need to subtract the \(2 x^{2}\) and 48 from the right side of the equation and put it on the left side. The new equation will read

apologiabiology

set to zero both sides since if we have
xy=0 then assume x and y=0
so

5x^2-27=2x^2+48
minus 2x^2+48 from both sides
3x^2-75=0
factor
3(x^2-25)=0
remember diffrence fo 2 perpfect squares (a^2-b^2=(a-b)(a+b))

3(x-5)(x+5)=0
set to zero
x-5=0
x=5

x+5=0
x=-5


x=5 and -5

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