To solve: 3y^{2}-y-14=0

Step 1
Determine the quadratic equations

Step 1
Given: ${\displaystyle 3y^2-y-14=0}$
All equations of the form ${\displaystyle a{x}^2+bx+c=0}$ can be solved using the quadratic formula: ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$. The quadratic formula gives two solutions, one when ${\displaystyle \pm }$ is addition and one when it is subtraction.
This equation is in standard form: ${\displaystyle a{x}^2+bx+c=0}$. Substitute 3 for a, -1 for b, and -14 for c in the quadratic formula, ${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$
${\displaystyle y=\frac{-(-1)\pm \sqrt{1-4\times 3(-14)}}{2\times 3}}$
Multiply -4 times 3.
${\displaystyle y=\frac{-(-1)\pm \sqrt{1-12(-14)}}{2\times 3}}$
Multiply -12 times -14.
${\displaystyle y=\frac{-(-1)\pm \sqrt{1+168}}{2\times 3}}$
Add 1 to 168.
${\displaystyle y=\frac{-(-1)\pm \sqrt{169}}{2\times 3}}$
Take the square root of 169.
${\displaystyle y=\frac{-(-1)\pm 13}{2\times 3}}$
The opposite of -1 is 1.
${\displaystyle y=\frac{1\pm 13}{2\times 3}}$
Multiply 2 times 3.
${\displaystyle y=\frac{1\pm 13}{6}}$
Now solve the equation ${\displaystyle y=\frac{1\pm 13}{6}}$ when ${\displaystyle \pm }$ is plus. Add 1 to 13.
${\displaystyle y=\frac{14}{6}}$
Reduce the fraction ${\displaystyle \frac{14}{6}}$ to lowest terms by extracting and canceling out 2
${\displaystyle y=\frac{7}{3}}$
Now solve the equation $y=\frac{1\pm 13}{6}$ when ${\displaystyle \pm }$ is minus. Subtract 13 from 1.
${\displaystyle y=\frac{-12}{6}}$
Divide -12 by 6
${\displaystyle y=-2}$
The equation is now solved
${\displaystyle y=\frac{7}{3}}$
${\displaystyle y=-2}$

Step 1
Split the middle term -y as 6y and -7y
Replace -y with +6y-7y in $3y^2-y-14=0$
$3y^2-7y+6y-14=0$
The above equation can be written as $3y\times y-7\times y+2\times 3y-2\times 7$
Apply distribtive property for first two terms and last two terms,
$3y\times y-7\times y+2\times 3y-2\times 7=y(3y-7)+2(3y-7)$
Again apply distributive property for y(3y-7)+2(3y-7)
y(3y-7)+2(3y-7)=(3y-7)(y+2)
So the factors are (3y-7) and (y+2)
$3y^2-y-14=(3y-7)(y+2)$
To find the value of y,
Take 3y-7=0
Add 7 on both sides,
3y-7+7=0+7
3y=7
Divide by 3 on both sides,
$\frac{3y}{3}=\frac{7}{3}$
$y=\frac{7}{3}$
Take y+2=0
Subtract 2 on both sides,
y-2+2=0-2
y=-2
The factors are $y=$$\frac{7}{3}$ and y=-2
So, the values of y are $\frac{7}{3}$ and -2