# How do you solve 2x^2-9x-5=0 by factoring?

Oct 5, 2015

The solutions are
color(blue)(x=-1/2, x=5

#### Explanation:

 2x^2−9x−5=0

We can Split the Middle Term of this expression to factorise it and thereby find the solutions.

In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 2 \cdot - 5 = - 10$

AND

${N}_{1} + {N}_{2} = b = - 9$

After trying out a few numbers we get ${N}_{1} = - 10$ and ${N}_{2} = 1$

$- 10 \cdot 1 = - 10$, and $1 + \left(- 10\right) = - 9$

 2x^2−color(blue)(9x)−5= 2x^2−color(blue)(10x+1x)−5

=2x(x-5)+ 1(x−5)

color(blue)((2x+1) (x-5) is the factorised form of the expression.

We now equate the factors to zero to obtain the solutions:

2x+1=0, color(blue)(x=-1/2
x-5=0, color(blue)(x=5