# How do you solve 7n ^ { 2} + 3n - 2= 0?

Mar 11, 2018

7n^2+3n−2=0
Step 1: Use quadratic formula with a=7, b=3, c=-2.
n= (−b±sqrt(b^2−4ac)/(2a))
n= (−(3)±sqrt((3)^2−4(7)(−2)))/(2(7)
n=(−3+sqrt65)/(14) or n=(−3-sqrt65)/(14)

#### Explanation:

Mar 11, 2018

n= (−3+ sqrt(65))/(14) and n= (−3- sqrt(65))/(14)

#### Explanation:

You can plug this into the quadratic formula in order to solve for $n$

$a x + b x + c = 0$

x= (−b± sqrt(b^2 −4ac))/(2a)

$7 {n}^{2} + 3 n - 2 = 0$

n= (−3± sqrt(3^2 −4(7)(-2)))/(2(7))

n= (−3± sqrt(9−(-56)))/(14)

n= (−3± sqrt(65))/(14)

n= (−3+ sqrt(65))/(14) and n= (−3- sqrt(65))/(14)

Remember to check your answer by plugging both answers into the original equation and see if the solutions match

If you checked your answer, you will see that they both work as solutions.

Mar 11, 2018

${n}_{1} = \frac{- 3 + \sqrt{65}}{14} = 0.3615898392$

${n}_{2} = \frac{- 3 - \sqrt{65}}{14} = - 0.7901612677$

#### Explanation:

Use the quadratic formula, $n = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the values of the quadratic equation into the form above from the current form, $a {n}^{2} + b n + c = 0$

$n = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 7 \cdot \left(- 2\right)}}{2 \cdot 7}$

Simplify

$n = \frac{- 3 \pm \sqrt{65}}{14}$

Solve for the solutions of $x \left(n\right)$

${n}_{1} = \frac{- 3 + \sqrt{65}}{14} = 0.3615898392$

${n}_{2} = \frac{- 3 - \sqrt{65}}{14} = - 0.7901612677$