Question

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

Answer 1

`5x=6x^2-3\ \ \ |-5x\\\\6x^2-5x-3=0\\\\a=6;\ b=-5;\ c=-3\\\\b^2-4ac=(-5)^2-4\cdot6\cdot(-3)=25+72=97\\\\x_1=(-b-√(b^2-4ac))/(2a);\ x_2=(-b+√(b^2-4ac))/(2a)\\\\x_1=(-(-5)-√(97))/(2\cdot6)=(5-√(97))/(12)\\\\x_2=(-(-5)+√(97))/(2\cdot6)=(5+√(97))/(12)`

Answer 2

`x=(5+√(97))/(12)` and
`x=(5-√(97))/(12)`

Explanation:

Using the quadratic formula to solve
`5x=6x^2-3`

To apply quadratic formula we make 0 on one side of the equation

`5x=6x^2-3`

Subtract 5x on both sides

`0=6x^2-5x-3`

Now apply quadratic formula. a=6, b=-5 and c=-3

`x=(-b+-√(b^2-4ac))/(2a)`

Plug in all the values in the formula

`x=(5+-√((-5)^2-4(6)(-3)))/(2(6))`

`x=(5+-√(97))/(12)`

`x=(5+√(97))/(12)` and
`x=(5-√(97))/(12)`