Question

How do you solve `2x^2+4x-5=0`?

Answer 1

Use the quadratic formula to find:

`x = -1+-sqrt(14)/2`

Explanation:

`2x^2+4x-5` is in the form `ax^2+bx+c` with `a=2`, `b=4` and `c=-5`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 4^2-(4*2*(-5)) = 16+40 = 56 = 2^2*14`

This is positive, but not a perfect square. So our quadratic equation has Real irrational roots.

We can find the roots using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

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

`=(-4+-sqrt(56))/(2*2)`

`=(-4+-2sqrt(14))/4`

`=-1+-sqrt(14)/2`

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Alternative Method

Multiply through by `2` first to make the leading term a perfect square, complete the squre, then factorise using the difference of square identity:

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

with `a=(2x+2)` and `b=sqrt(14)` as follows:

`0 = 4x^2+8x-10`

`=(2x+2)^2-4-10`

`=(2x+2)^2-(sqrt(14))^2`

`=((2x+2)-sqrt(14))((2x+2)+sqrt(14))`

`=(2x+2-sqrt(14))(2x+2+sqrt(14))`

`=(2(x+1-sqrt(14)/2))(2(x+1+sqrt(14)/2))`

`=4(x+1-sqrt(14)/2)(x+1+sqrt(14)/2)`

Hence `x = -1+-sqrt(14)/2`