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16x^2=9 using ax^2+bx+c=0 factor the left hand side of the equation into two linear factors

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RedRicinid
[tex]16x^2=9[/tex] becomes [tex]16x^2-9=0[/tex] ([tex]+0x[/tex] is implied)

To solve a quadratic like this we need to find two numbers that multiply to equal ac and add up to b. In this case, two numbers that multiply to equal -144 and add to equal 0. These two numbers are 12 and -12. Factor using these numbers.

[tex](x+12)(x-12)=0[/tex]

And if you ever need to find x you would just find any value that would cause one of the two to equal 0. Since 0 times anything is still 0, this would work. These values would be 12 and -12. (12-12 = 0, -12+12 = 0)
Panoyinid
16x² = 9
16x² - 9 = 0
16x² + 0x - 9 = 0
x = 0 +/- √(0² - 4(16)(-9))
                  2(16)
x = 0 +/- √(0 + 576)
                 32
x = 0 +/- √(576)
              32
x = 0 +/- 24
           32
x = 0 + 0.75
x = 0 + 0.75      x = 0 - 0.75
x = 0.75            x = -0.75