Question

Solve each equation using the quadratic formula. Find the exact solutions.

6n^2 + 4n = -11

Answer 1

Answer:

$n_1=-\frac{1}{3}+1.31i\\n_2=-\frac{1}{3}-1.31i$

Step-by-step explanation:

To solve this problem you  must apply the proccedure shown below:

1. You have that the quadratic formula is:

$n=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}$

2. To solve the quadratic equation you must substitute the values, you have that:

$a=6\\b=4\\c=11$

Then:

 $n=\frac{-4+/-\sqrt{4^{2}-4(6)(11)}}{2(6)}$

3, Therefore, you obtain the following result:

$n_1=-\frac{1}{3}+1.31i\\n_2=-\frac{1}{3}-1.31i$

Answer 2

Answer:

(-2±√62i ) / 12

Step-by-step explanation:

Given equation is :

6n²+4n =-11

Adding 11 to both sides of above equation, we get

6n²+4n+11=-11+11

6n²+4n+11= 0

an²+bn+c = 0 is general quadratic equation.

n =(-b±√b²-4ac) / 2a is solution of general equation.

Comparing  general quadratic equation with given quadratic equation,we get

a = 6, b = 4 and c = 11

Putting above values in quadratic formula,we get

n= (-4±√4²-4(6)(11))/ 2(6)

n = ( -4±√16-264) / 12

n = (-4±√-248) / 12

n = ( -4±√-1√248) / 12

n = (-4±√4×62i) / 12

n = (-4± 2√62i) / 12

n = 2(-2±√62i) / 12

n = (-2±√62i ) / 12 is solution of given equation.