Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction

Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?

5x = 6x^2 – 3

Subtract 5x from both sides:

0 = 6x^2 – 5x – 3

a = 6, b = -5, c = -3

x = (-b ± √(b^2 - 4ac))/(2a)

x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))

x = (5 ± √(25 + 72))/12

x = (5 ± √(97))/12

Step-by-step explanation:

The answer is B

Step-by-step explanation:

6 meters

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)