Question

N² + 14 = 12nhow do I solve by quadratic formula ​

Answer 1

Answer:

Step-by-step explanation:

The quadratic formula is $\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

Ignore the weird A at the beginning, I don't know why it is there.

To get your equation into a quadratic equation, we have to move 12n to the other side, giving us

$n^{2} -12n+14$

So in this case, our a=1, b=-12, and c=14. Remember $ax^{2}+bx+c$

So we plug these values into our formula

$\frac{12±\sqrt{144-4(14)} }{2}$. Again, ignore the weird A.

simplify and you will get

$\frac{12±\sqrt{88} }{2}$

simplify the square root and you get $2\sqrt{22}$

divide the $2\sqrt{22}$ and $12$ by the $2$ on the bottom and you will get $\sqrt{22}$ and $6$

So your answers are $6-\sqrt{22}$ and $6+\sqrt{22}$

Answer 2

Answer:

n = 6 + $\sqrt{22}$  or  n = 6 - $\sqrt{22}$

Step-by-step explanation:

We can solve this equation using the quadratic formula OR Completing the Square method.

n² + 14 = 12n

rearrange :  n² - 12n + 14  = 0  

here  a= 1 , b = -12,  c = 14

the quadratic formula says:   x =  - b/ (2a)  +  root(b^2 - 4ac) / (2a)

or  x =  - b/ (2a)  -  root(b^2 - 4ac) / (2a)

x =  - (-12)/ (2)  +  root((-12)^2 - 4*14) / (2)

x = 6  +  root (144 - 56) / 2

x = 6 + root(88)/2

x = 6 + root(4*22) / 2

x = 6 + 2*root(22)/2

x = 6 + root(22)  = 6 + $\sqrt{22}$

so   x =6 + $\sqrt{22}$   or  x = 6 - $\sqrt{22}$

In this case  x = n

n = 6 + $\sqrt{22}$  or  n = 6 - $\sqrt{22}$