 Roger Smith

2021-12-28

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

Expert

Step 1 chumants6g

Expert

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

Expert

Step 1
Split the middle term -y as 6y and -7y
Replace -y with +6y-7y in $3{y}^{2}-y-14=0$
$3{y}^{2}-7y+6y-14=0$
The above equation can be written as $3y×y-7×y+2×3y-2×7$
Apply distribtive property for first two terms and last two terms,
$3y×y-7×y+2×3y-2×7=y\left(3y-7\right)+2\left(3y-7\right)$
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)
$3{y}^{2}-y-14=\left(3y-7\right)\left(y+2\right)$
To find the value of y,
Take 3y-7=0
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

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