Question

How do you solve `2x^2 -: 3x = 14` ?

Answer 1

See explanation...

Explanation:

I think the question should have had subtraction rather than division, but I will answer both possibilities:

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How to solve `bb((2x^2)/(3x)=14)`

Multiply both sides by `3x` to get:

`2x^2 = 42x`

Divide both sides by `2x` to find:

`x = 21`

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How to solve `bb(2x^2-3x=14)`

Subtract `14` from both sides to get:

`2x^2-3x-14 = 0`

Use an AC method to factorise:

Look for a pair of factors of `AC=2*14=28` which differ by `B=3`.

The pair `7, 4` works.

Use this to split the middle term then factor by grouping:

`0 = 2x^2-3x-14`

`=2x^2-7x+4x-14`

`=(2x^2-7x)+(4x-14)`

`=x(2x-7)+2(2x-7)`

`=(x+2)(2x-7)`

Hence `x=-2` or `x=7/2`