# How do you solve  2x^2 - 7x - 9 = 0?

Jul 2, 2015

I found:
${x}_{1} = - 1$
${x}_{2} = \frac{9}{2}$

#### Explanation:

We can use the Quadratic Formula but...it is boring!
Divide by $2$:
${x}^{2} - \frac{7}{2} x - \frac{9}{2} = 0$
Rearrange:
${x}^{2} - \frac{7}{2} x = \frac{9}{2}$
Add ans subtract $\frac{49}{16}$;
${x}^{2} - \frac{7}{2} x + \frac{49}{16} - \frac{49}{16} = \frac{9}{2}$
Rearrange:
${x}^{2} - \frac{7}{2} x + \frac{49}{16} = \frac{9}{2} + \frac{49}{16}$
${\left(x - \frac{7}{4}\right)}^{2} = \frac{72 + 49}{16} = \frac{121}{16}$
$x - \frac{7}{4} = \pm \sqrt{\frac{121}{16}} = \pm \frac{11}{4}$
So:
${x}_{1} = \frac{7}{4} - \frac{11}{4} = - 1$
${x}_{2} = \frac{7}{4} + \frac{11}{4} = - \frac{18}{4} = \frac{9}{2}$

Jul 2, 2015

Solve $y = 2 {x}^{2} - 7 x - 9 = 0$

#### Explanation:

To solve this type of quadratic equation, use shortcut. It can save you a lot of time and work.
(a - b + c = 0) --> 2 real roots: x = - 1, and $x = - \frac{c}{a} = \frac{9}{2}$

Remind of Shortcut:

1. When (a + b + c = 0) -> 2 real roots: 1 and $\left(\frac{c}{a}\right)$

2. When (a - b + c = 0 -> 2 real roots: - 1 and $\left(- \frac{c}{a}\right)$