# 6x^2-10x-16=0?

Jun 15, 2018

Solve the quadratic to get $x = - 1 , \frac{8}{3}$

#### Explanation:

I assume the question wants the solutions for $x$.

Divide through to make the first coefficient 1:
${x}^{2} - \frac{5}{3} x - \frac{8}{3} = 0$

Complete the square:
${\left(x - \frac{5}{6}\right)}^{2} - \frac{25}{36} - \frac{8}{3} = 0$
${\left(x - \frac{5}{6}\right)}^{2} = \frac{121}{36}$

$x = \frac{5}{6} \pm \sqrt{\frac{121}{36}} = \frac{5}{6} \pm \frac{11}{6}$

So $x = - 1 , \frac{8}{3}$.

Jun 16, 2018

$y = 6 {x}^{2} - 10 x - 16 = 0$
Since a - b + c = 0, use shortcut.
The 2 real roots are:
x = -1 ,and $x = - \frac{c}{a} = \frac{16}{6} = \frac{8}{3}$

Reminder of shortcut .
If a + b + c = 0, the 2 real roots are: x = 1, and $x = \frac{c}{a}$
Example 1. Solve: $3 {x}^{2} + 7 x - 10 = 0$
x = 1, and $x = \frac{c}{a} = - \frac{10}{3}$
If a - b + c = 0, the 2 real roots are: x = -1, and $x = - \frac{c}{a}$
Example 2. Solve: $4 {x}^{2} - 5 x - 9 = 0$
x = -1, and $x = - \frac{c}{a} = \frac{9}{4}$

Jun 16, 2018

Another way

#### Explanation:

if $a {x}^{2} + b x + c = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case $a = 6 , b = - 10 , c = - 16$

$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - \left(4 \cdot 6 \cdot \left(- 16\right)\right)}}{2 \cdot 6}$

$x = \frac{10 \pm \sqrt{484}}{12}$

$x = \frac{10 \pm 22}{12}$

$x = \left\{\frac{8}{3} , - 1\right\}$