# How do you factor 6t^2-17t+12=0?

Sep 16, 2015

color(blue)((3t-4)(2t - 3)

#### Explanation:

$6 {t}^{2} - 17 t + 12 = 0$

We can Split the Middle Term of this expression to factorise it.

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

${N}_{1} \cdot {N}_{2} = a \cdot c = 6 \cdot 12 = 72$
and
${N}_{1} + {N}_{2} = b = - 17$

After trying out a few numbers we get ${N}_{1} = - 9$ and ${N}_{2} = - 8$
$\left(- 9\right) \cdot \left(- 8\right) = 72$, and $- 9 - 8 = - 17$

$6 {t}^{2} - 17 t + 12 = 0$

$6 {t}^{2} - 9 t - 8 t + 12 = 0$

$3 t \left(2 t - 3\right) - 4 \left(2 t - 3\right) = 0$

color(blue)((3t-4)(2t - 3) is the factorised form of the expression.