Help with solving using quadratic formula??

Question

x^2-3x = 4x - 1
This one has me stumped. Can someone explain???

Answer 1

x^2 - 3x = 4x - 1
x^2 - 7x + 1 = 0


Remember, for ax^2 + bx + c = 0,
x = [-b +- sqrt(b^2 - 4ac)]/[2a]


So for here,
x = [7 +- sqrt( (-7)^2 - 4(1)(1) )] / [2]
x = 7/2 +- 3*sqrt(5) / 2


x = 6.85 or 0.146 (approx)

Answer 2

x^2-3x = 4x - 1

Group all the terms on the left side by subtracting 4x and adding 1 to both sides

x² - 7x +1= 0

ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a

Now enter coefficients into quadratic formula

x = (7 ± √(49 - 4))/2

x = (7 ± √(45))/2

x = (7 ± 3√(5))/2

x = 6.854, x= 0.1459
.

Answer 3

Just bring the 4x and the -1 to the other side of the equal sign. Remember when you move a number to the other side of the equal sign the (-) or (+) sign changes.
so your equation becomes x^2-7x+1=o
then use the quadratic formula

Answer 4

x^2-3x=4x-1 is x^2-7x+1=0
the general solution for x in the quadratic equation a(x^2)+b(x)+c=0 is
x = {-b[+(or)-]sqrt(b^2-4ac)}/2a
a=coefficient of x^2
b=coefficient of x
c=constant
we have x^2-7x+1=0
a=1, b= -7, c=1
x= {7[+(or)-]sqrt(49-4(1))}/2
x= {7[+(or)-]sqrt(45)}/2
x={7[+(or)-]3*sqrt(5)}2
note1:
since 45 = 9*5, sqrt(45)= sqrt(9*5)= sqrt([3^2]*5)=3*sqrt(5)
note2:
since we have a quadratic equation of x, ie a equation of x with degree 2.
so the final solution is,
x=(7+[3*sqrt(5)])/2 and x=(7-[3*sqrt(5)])/2
where sqrt(5) = 2.236
x = (7+[3*2.236])/2 and x = (7-[3*2.236])/2
x=13.708/2 and x= 0.292/2
we get x = 6.854 and x = 0.146 are the required solution for the given equation

Answer 5

Get it in standard form (one side is zero).

x^2 -7x +1 =0

a=1 b=-7 c=1 Substitute these into the formula

x= [-b±√(b^2 - 4ac)]/2a

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Answer 6

x² - 7x + 1 = 0
x = [ 7 ± √(49 - 4) ] / 2
x = [ 7 ± √45 ] / 2
x = 6.85 , x = 0.146