# How do you solve h^2+48=16h?

Mar 17, 2018

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{16 h}$ from each side of the equation to put the equation in standard quadratic form:

${h}^{2} - \textcolor{red}{16 h} + 48 = 16 h - \textcolor{red}{16 h}$

${h}^{2} - 16 h + 48 = 0$

Now, we can use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 16}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{48}$ for $\textcolor{g r e e n}{c}$ gives:

$h = \frac{- \textcolor{b l u e}{- 16} \pm \sqrt{{\textcolor{b l u e}{- 16}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{48}\right)}}{2 \cdot \textcolor{red}{1}}$

$h = \frac{16 \pm \sqrt{256 - 192}}{2}$

$h = \frac{16 \pm \sqrt{64}}{2}$

$h = \frac{16 - 8}{2}$ and $h = \frac{16 + 8}{2}$

$h = \frac{8}{2}$ and $h = \frac{24}{2}$

$h = 4$ and $h = 12$

The Solution Set Is: $h = \left\{4 , 12\right\}$

Another method is to factor the quadratic as:

$\left(h - 12\right) \left(h - 4\right) = 0$

The solve each term on the left for $0$:

Solution 1:

$h - 12 = 0$

$h - 12 + \textcolor{red}{12} = 0 + \textcolor{red}{12}$

$h - 0 = 12$

$h = 12$

Solution 2:

$h - 4 = 0$

$h - 4 + \textcolor{red}{4} = 0 + \textcolor{red}{4}$

$h - 0 = 4$

$h = 4$

The Solution Set Is: $h = \left\{4 , 12\right\}$

Mar 17, 2018

$h = 4 \text{ or } h = 12$

#### Explanation:

$\text{rearrange equation into "color(blue)"standard form}$

$\Rightarrow {h}^{2} - 16 h + 48 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{the factors of + 48 which sum to - 16 are - 12 and - 4}$

$\Rightarrow \left(h - 12\right) \left(h - 4\right) = 0$

$\text{equate each factor to zero and solve for h}$

$h - 12 = 0 \Rightarrow h = 12$

$h - 4 = 0 \Rightarrow h = 4$

$\Rightarrow h = 12 \text{ or "h=4" are the solutions}$