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Question
`7x^2+3x-4=0`
Solution
`7x^2+3x-4=0`
⇒`49x^2+21x-28=0` (Multiplying both sides by 7)
⇒`49x^2+21x=28`
⇒`(7x)^2+xx7x xx3/2+(3/2)^2=28+(3/2)^2` [Adding (3/2)^2 on both sides]
⇒`(7x+3/2)^2=28+9/4=121/4=(11/2)^2`
⇒`7x+3/2=+-11/2` (Taking square root on both sides)
⇒`7x+3/2=11/2 or 7x+3/2=-11/2`
⇒`7x=11/2-3/2=8/2=4 or 7x=11/2-3/2=-14/2=-7`
⇒`x=4/7 or x=-1`
Hence, `4/7 and -1` are the roots of the given equation.
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