Quadratic formula, help?

Question

x^2 + 6x = 4?

Thanks

Answer 1

x = (-6 +-sq.root of 6^2-(4)(-4))/2

Answer 2

rearrange the equation as, x^2 + 6x - 4 = 0

compare with the general form as follows,

(a) x^2 + (b) x + c = 0

which gives you, a = 1, b = 6 and c = -4

the quadratic formula for finding roots of x is,

x = [-b +/- sqrt (b^2-4ac)] / [2*a]

x = [-6 +/- sqrt (36-4(1)(-4))]/[2(1)]

x =[ -6 +/- sqrt 52 ]/2

x = -3 +/- [sqrt (13*4)]/2

x= -3 +/- [2sqrt13/2]

x = -3+/-sqrt(13)

Answer 3

x squared +6x-4=0
a=1,b=6,c=-4.
The quadratic formula is x=-b plus minus b squared-- 4 times a times c[all of this in the square root sign. For example, in this case,c it'd be 36-16=20. The square root of that.] divided by 2a.

Answer 4

This is not a perfect square, so if you use the quadratic formula, you should get

x = -3 (+/-) [ squart_root(20) / 2 ]

simplifying

x = -3 (+/-) squart_root(5)

Answer 5

x^2 + 6x - 4 = 0

if the equation is ax^2 + bx + c = 0,

formula to find the roots is:

(- b (+-) sqroot (b^2-4ac)) / 2a

apply the formula

first root is

(- 6 + sqroot (36-16)) / 2 = -3 + 2 sqroot 5

second root is

-3 - 2 sqroot 5

Answer 6

x^2+6x-4=0

a=1 b=6 c=-4

-6+/- sqrt 6^2-4(1)(-4)/2
-6+/- 36+16/2
-6+/- 52/2
-6 +/- 13*4/2
-3+/-sqrt13

-3 +/- sqrt13

Answer 7

x^2 + 6x -4=0

a=1
b=6
c=-4

x1=-3-sqrt(13)
x2=-3+sqrt(13)

Answer 8

x has two values

x1 = - 3+(13)^(1/2)
x2 = - 3 - (13)^(1/2)

GENERAL FORMULA IS

FOR a(x)^2 + bx +c = 0

x1 = (-b) + ( (b)^2 - 4ac)^(1/2)
x2 = (-b) - ((b)^2 - 4ac)^(1/2)