# How do you solve 6h^2+17h+12=0?

May 20, 2015

$6 {h}^{2} + 17 h + 12 = 0$

We can Split the Middle Term of this equation to factorise it and then find the solution for the equation

In this technique, if we have to factorise an expression like $a {h}^{2} + b h + c$, we need to think of 2 numbers such that:
${N}_{1} \times {N}_{2} = a \times c = 6 \times 12 = 72$
and
${N}_{1} + {N}_{2} = b = 17$

After trying out a few numbers we get ${N}_{1} = 9$ and ${N}_{2} = 8$
$9 \times 8 = 72$, and $9 + 8 = 17$

$6 {h}^{2} + 17 h + 12 = 6 {h}^{2} + 9 h + 8 h + 12$
$= 3 h \left(2 h + 3\right) + 4 \left(2 h + 3\right)$
$= \left(3 h + 4\right) \left(2 h + 3\right)$
the solution for the equation is :
$\textcolor{b l u e}{h} = - \frac{4}{3}$
$\textcolor{b l u e}{h} = - \frac{3}{2}$

May 20, 2015

The answer is $h = - \frac{3}{2} , - \frac{4}{3}$

Problem: Solve $6 {h}^{2} + 17 h + 12 = 0$ .

Multiply $6$ times $12$ to get $72$.
Find two numbers that when multiplied equal $72$ and when added equal $17$.
Numbers $8$ and $9$ meet the criteria.
Write the equation substituting $8 h$ and $9 h$ for $17 h$.

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

Separate the first two terms from the second two terms.

$\left(6 {h}^{2} + 8 h\right) + \left(9 h + 12\right)$

Factor out the GCF from both sets of terms.

$2 h \left(3 h + 4\right) + 3 \left(3 h + 4\right) = 0$

Factor out the GCF $\left(3 h + 4\right)$ .

$\left(3 h + 4\right) \left(2 h + 3\right) = 0$

Solve for $h$.

$3 h + 4 = 0$

$3 h = - 4$

$h = - \frac{4}{3}$

$2 h + 3 = 0$

$2 h = - 3$

$h = - \frac{3}{2}$

$h = - \frac{3}{2} , - \frac{4}{3}$