Find the zeroes of the quadratic equation 4x^-7?
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4x²-7=(2x)²-(√7)19 =(2x-√7)(2x+√7)
2x-√7=0 and 2x+√7=0
x=√7/2 and x= -√7/2
x=√7/2,-√7/2
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2x-√7=0 and 2x+√7=0
x=√7/2 and x= -√7/2
x=√7/2,-√7/2
hope it helps you!
please mark me as brainliest if you are satisfied
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