For this case, we have the following equation of the second degree:
[tex]30x ^ 2-28x + 6 = 0[/tex]
If we divide between 2 on both sides of the equation, we will have:
[tex]15x ^ 2-14x + 3 = 0[/tex]
Where:
[tex]a = 15\\b = -14\\c = 3[/tex]
The solutions will come from:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
Substituting:
[tex]x = \frac {- (- 14) \pm \sqrt {(- 14) ^ 2-4 (15) (3)}} {2 (15)}\\x = \frac {14 \pm \sqrt {196-180}} {30}\\x = \frac {14 \pm \sqrt {16}} {30}\\x = \frac {14 \pm4} {30}[/tex]
So, we have:
[tex]x_ {1} = \frac{14 + 4} {30} = \frac {18} {30} = \frac {18} {30} = \frac {3} {5}\\x_ {2} = \frac {14-4} {30} = \frac {10} {30} = \frac {1} {3}\\[/tex]
Answer:
[tex]x_ {1} = \frac {3} {5}\\x_ {2} = \frac {1} {3}[/tex]
Option A
Option E
