Question 145098
What is the correct substitution in the quadratic formula for the solution of the following
8x^2+3a^2=10ax- 
Rearrange to get:
8x^2 - 10ax + 3a^2
Quadratic with a = 8 ; b = -10a ; c = 3a^2
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Using the quadratic formula you would get:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
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Substitute the values of a, b, c to get:
x = [-10a +- sqrt((10a)^2 - 4(8)(3a^2))]/(16)
x = [-10a +- sqrt(100a^2 - 96a^2)]/16
etc.
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A. x=-10a+-(100a^2-4*3a^2*10)^(1/2)
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2*8 
B. x= -3a^2+-(9a^2-4*8+10a)^(1/2)
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2*8 
c. x=10+-(100-4*8*3)^(1/2)
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2*8 
d. x=10a+-(100a^2-4*8*3a^2)

Cheers,
Stan H.