solve the following quadratic equations
1.300/x-300/x+10=1
2.x-1/x=1
3.x^{2} +[3x+3]^{2} =[3x+4]^{2]
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Step-by-step explanation:
1. given equation is not a quadratic equation
2. x-1/x=1
= x^2-1/x=1
= x^2-1=x
= x^2-x-1=0
the answer is in the form of ax^2+bx+c=0
answer is x^2-x-1=0
3. given
x^2+(3x+3)^2=(3x+4)^2
x^2+ 9x^2+9+18x=9x^2+16+24x
x^2-6x-7=0
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