Jan 18, 2018

See below

Explanation:

The quadratic formula is $x = \frac{- b \pm \sqrt{D}}{2 a}$
Here $D = {b}^{2} - 4 a c$
Only to need to put the values in the formula.
a = 6
b = 5
c = -6

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(6\right) \left(- 6\right)}}{2 \cdot 6}$
$x = \frac{- 5 \pm \sqrt{25 + 144}}{12}$
$x = \frac{- 5 \pm \sqrt{169}}{12}$
$x = \frac{- 5 \pm \left(13\right)}{12}$
So x is either,
$\frac{- 5 - 13}{12}$
=$- \frac{18}{12}$
=$- \frac{3}{2}$
Or
$\frac{- 5 + 13}{12}$
=$\frac{8}{12}$
=$\frac{2}{3}$
Hope it helps you

Jan 18, 2018

See explanation.

Explanation:

1) $f \left(x\right) = 6 {x}^{2} + 5 x - 6$
$= 6 {x}^{2} + 9 x - 4 x - 6$
$= 3 x \left(2 x + 3\right) - 2 \left(2 x + 3\right)$
$= \left(2 x + 3\right) \left(3 x - 2\right)$

That's it for part1

2)
$f \left(x\right) = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Here, a=6, b=5, c=-6
Plugging in the values, the roots of the equation will be:
(-5+- sqrt(5^2-4*6*(-6)))/(2*6
Simplify the equation, and the roots will be
$\frac{- 5 \pm \sqrt{169}}{12}$
$= \frac{- 5 + \sqrt{169}}{12} \mathmr{and} \frac{- 5 - \sqrt{169}}{12}$
$= \frac{- 5 + 13}{12} \mathmr{and} \frac{- 5 - 13}{12}$
$= \frac{8}{12} \mathmr{and} - \frac{18}{12}$
$= \frac{2}{3} \mathmr{and} - \frac{3}{2}$

therefore, the equation will be:

$\left(x - \frac{2}{3}\right) \left(x + \frac{3}{2}\right) = 0$

Thus, your final equation will be:
$\left(2 x + 3\right) \left(3 x - 2\right)$

$T h a n k s .$
Hope you got it.

Jan 18, 2018

Factoring Method

color(blue)(f(x) = 6x^2+5x-6=(3x-2)(2x+3)

color(blue)(x = 2/3, x= -3/2

Explanation:

Given:

color(green)(f(x) = 6x^2+5x-6

The Standard Form of a Quadratic Equation:

color(red)(y = f(x) = ax^2+bx+c = 0

From our problem:

a = 6; b = 5; and c = -6

$\textcolor{b r o w n}{M e t h o d .1} \text{ }$Factoring Method

Using the Standard Form

$y = f \left(x\right) = a {x}^{2} + b x + c$

we find $\textcolor{b l u e}{u}$ and $\textcolor{b l u e}{v}$ such that

color(green)(u *v = a*c and u + v = b

Then we need to group them as shown below:

$a {x}^{2} + u x + v x + c$

We have

color(green)(f(x) = 6x^2+5x-6=0

we find $\textcolor{b l u e}{u}$ and $\textcolor{b l u e}{v}$ as:

color(green)(u = [-4] and v = [9]

So, the middle term $\textcolor{b l u e}{5 x}$ can be written as color(blue)([-4x+9x]

We can now write our $f \left(x\right)$ as

color(green)(f(x) = 6x^2-4x+9x-6=0

$\Rightarrow 6 {x}^{2} - 4 x + 9 x - 6 = 0$

$\Rightarrow 2 x \left(3 x - 2\right) + 3 \left(3 x - 2\right) = 0$

$\Rightarrow \left(3 x - 2\right) \left(2 x + 3\right) = 0$

We get

$\left(3 x - 2\right) = 0 , \left(2 x + 3\right) = 0$

$3 x - 2 \Rightarrow 3 x = 2$ hence $x = \frac{2}{3}$

$2 x + 3 = 0 \Rightarrow 2 x = - 3$ hence $x = - \frac{3}{2}$

Hence, $\textcolor{b l u e}{x = \frac{2}{3} , x = - \frac{3}{2}}$

$\textcolor{b r o w n}{M e t h o d .2} \text{ }$Using Quadratic Formula

color(blue)(x = [-b +- sqrt(b^2 - 4ac)]/(2a)

From our problem:

a = 6; b = 5; and c = -6

Substituting these values of $a , b \mathmr{and} c$ in our formula

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot 6 \cdot \left(- 6\right)}}{2 \cdot 6}$

$\Rightarrow \frac{- 5 \pm \sqrt{25 + 144}}{12}$

$\Rightarrow \frac{- 5 \pm \sqrt{169}}{12}$

$\Rightarrow \frac{- 5 \pm 13}{12}$

Hence,

$x = \frac{- 5 + 13}{12} , x = \frac{- 5 - 13}{12}$

$x = \frac{8}{12} , x = - \frac{18}{12}$

$x = \frac{2}{3} , x = - \frac{3}{2}$

Hence, $\textcolor{b l u e}{x = \frac{2}{3} , x = - \frac{3}{2}}$

We can observe that both the methods yield the same values for $x$

Hope you find this solution helpful.