Question

How do you solve `3x^2-10x+8=0`?

Answer 1

The answer is `S={2, 4/3}`

Explanation:

We compare this equation to

`ax^2+bx+c=0`

`3x^2-10x+8=0`

The discriminant is

`Delta=b^2-4ac=100-4*3*8=100-96=4`

As, `Delta>0`, there are 2 real roots

`x=(-b+-sqrt(Delta))/(2a)`

`x_1=(10+2)/6=2`

`x_2=(10-2)/6=8/6=4/3`

Answer 2

`x=4/3` and`x=2`

Explanation:

`3x^2-10x+8=0`

`(3x-4)(x-2)`

`3x=4`
multiply both sides by`1/3`

`x=4/3`

`x-2=0`

`x=2`
substitute `x=4/3 and x=2`

`3(4/3)^2-10(4/3)+8=0`

`3 xx 16/9-40/3+8=0`

`cancel48^16/cancel9^3-40/3+8=0`

`5 1/3-13 1/3+8=0`

`-8+8=0`

`3(2)^2-10(2)+8=0`

`12-20+8=0`

`-8+8=0`