# How do you solve 2x^2 - x = 6 graphically and algebraically?

Mar 25, 2016

$x = 2 , \mathmr{and} - \frac{3}{2}$
$- \frac{3}{2} \text{can also be written as} - 1.5$

#### Explanation:

Let's draw graph of the expression using graphing tool.
graph{2x^2-x-6 [-9.86, 10.14, -7.64, 2.36]}
Observe in the graph where it intersects $x$ axis.
$\left(- 1.5 , 0\right) \mathmr{and} \left(2 , 0\right)$
$- 1.5 \mathmr{and} 2$ are solutions of the equation.

Given equation is $2 {x}^{2} - x = 6$.
Taking all terms to left hand side we get
$2 {x}^{2} - x - 6 = 0$
Let's try split the middle term method.
Product of first and third term $= - 12 {x}^{2}$
Let's find two parts of middle term so that their product is equal to above. $12$ has factors of $6 \times 2 , 4 \times 3 , 12 \times 1$. We see that $4 \times 3$ parts work for us. Splitting middle term as in two parts $- 4 x \mathmr{and} 3 x$ we obtain

$2 {x}^{2} - 4 x + 3 x - 6 = 0$, pair first two and last two
$\left(2 {x}^{2} - 4 x\right) + \left(3 x - 6\right) = 0$, Take out Common factor $2 x$ out of first pair and $3$ out of second pair.
$2 x \left(x - 2\right) + 3 \left(x - 2\right) = 0$, Take out Common factor $x - 2$ out of two
$\left(x - 2\right) \left(2 x + 3\right) = 0$, Set each factor equal to $0$
$\left(x - 2\right) = 0$ .......(1)
$\left(2 x + 3\right) = 0$ .....(2)
From (1) $x = 2$
From (2) $x = - \frac{3}{2}$