# need helpneed help please:

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We can also use ac method here.

a = 2, b = 9 and c = -5.

So, a*c = 2*-5 = -10.

We need to find two numbers which has a product of -10, and has a sum of 9.

Factors of -10 and their sums are the following:

-10 and 1 ===> -10 + 1 = -9

10 and -1 ===> 10 + (-1) = 9

5 and - 2 ===> 5 + (-2) = 3

-5 and 2 ===> -5 + 2 = -3

Hence, the two numbers we are looking for are 10 and -1.

We can rewrite the equation like this:

2x^2 + 10x - 1x - 5 = 0

Splitting the terms into two groups, where each group has a gcf.

(2x^2 + 10x) + (-1x - 5) = 0

Factoring out the gcf on each group.

2x(x + 5) - 1(x + 5) = 0

Factor out a common factor.

(x + 5)(2x - 1) = 0

Equating each factor to zero.

x + 5 = 0

x = -5

2x - 1 = 0

x = 1/2

Hence, **x = {-5, 1/2}**.

I suppose that you need help in finding the roots of the equation. I hope that the answer not to come too late.

You could apply quadratic formula for finding the roots of the equation:

x1 = [-b + sqrt(b^2 - 4ac)]/2a

x2 = [-b - sqrt(b^2 - 4ac)]/2a

All we have to do know is to identify the coefficients of the equation and to calculate the roots:

a = 2, b = 9 and c = -5

x1 = [-9 + sqrt(81 + 40)]/4

x1 = [-9+sqrt(121)]/4

x1 = (-9+11)/4

x1 = 2/4

x1 = 1/2

x2 = (-9-11)/4

x2 = -5

**So, the values of x that cancel the equation are x1 = 1/2 and x2 = -5.**

2x^2 +9x-5=0