# How do you solve 2y ^ { 2} - 5x - 12= 0?

Dec 5, 2016

$y = \pm \sqrt{\frac{5}{2} x + 6}$

#### Explanation:

Taking it 1 step at a time so that you can see better what is going on

Using first principles (short cut method taken from this approach)

$\textcolor{p u r p \le}{\text{Step 1}}$

Add $\textcolor{red}{5 x}$ to both sides

$\textcolor{b l u e}{2 {y}^{2} - 5 x - 12 = 0 \text{ "->" } 2 {y}^{2} \textcolor{red}{+ 5 x} - 5 x - 12 = 0 \textcolor{red}{+ 5 x}}$

$\text{ "2y^2" "+0" } - 12 = 5 x$

$\text{ } 2 {y}^{2} - 12 = 5 x$

...................................................................................................

$\textcolor{p u r p \le}{\text{Step 2}}$

Add $\textcolor{red}{12}$ to both sides

$\textcolor{b l u e}{2 {y}^{2} - 12 = 5 x \text{ "->" } 2 {y}^{2} \textcolor{red}{+ 12} - 12 = 5 x \textcolor{red}{+ 12}}$

$\text{ "2y^2" "+0" } = 5 x + 12$

$\text{ } 2 {y}^{2} = 5 x + 12$

...................................................................................................

$\textcolor{p u r p \le}{\text{Step 3}}$

Divide both sides by 2 (same as $\textcolor{red}{\times \frac{1}{2}}$)

color(blue)(2y^2=5x+12" "->" "2/(color(red)(2))y^2=5/(color(red)(2))x+12/(color(red)(2))

But $\frac{2}{2} = 1$ giving:$\text{ } {y}^{2} = \frac{5}{2} x + 6$

...................................................................................................

$\textcolor{p u r p \le}{\text{Step 4}}$

Square root both sides:$\text{ } \sqrt{{y}^{2}} = \sqrt{\frac{5}{2} x + 6}$

$\text{ } y = \pm \sqrt{\frac{5}{2} x + 6}$
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$\textcolor{p u r p \le}{\underline{\text{Foot note}}}$

Using an example $\left(- 2\right) \times \left(- 2\right) = + 4$
$\text{ } \left(+ 2\right) \times \left(+ 2\right) = + 4$

As both $\left(- 2\right) \times \left(- 2\right) \text{ and } \left(+ 2\right) \times \left(+ 2\right)$ both equal 4 then

$\sqrt{4} = \text{plus or minus } 2 \to \pm 2$

Which is why we end up with $y = \pm \sqrt{\frac{5}{2} x + 6}$