What are the solutions to the equation 120n^2-328n=-224 ?

The solutions to the equation 120n^2-328n=-224 are n= 7/5 ,n= 4/3

Find the zeros of 120n^2-328n=-224

The zeros of 120n^2-328n=-224 are n= 7/5 ,n= 4/3

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120 n^{2} - 328 n = - 224

n = \frac{7}{5}, n = \frac{4}{3}

Decimal Notation

n = 1.4, n = 1.33333 \dots

Solution steps

120 n^{2} - 328 n = - 224

Move

224

to the left side

120 n^{2} - 328 n + 224 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 328 ) \pm ( - 328 ) ^{2} - 4 \cdot 120 \cdot 224}{2 \cdot 120}

(- 328)^{2} - 4 \cdot 120 \cdot 224 = 8

n_{1, 2} = \frac{- ( - 328 ) \pm 8}{2 \cdot 120}

Separate the solutions

n_{1} = \frac{- ( - 328 ) + 8}{2 \cdot 120}, n_{2} = \frac{- ( - 328 ) - 8}{2 \cdot 120}

n = \frac{- ( - 328 ) + 8}{2 \cdot 120} : \frac{7}{5}

n = \frac{- ( - 328 ) - 8}{2 \cdot 120} : \frac{4}{3}

The solutions to the quadratic equation are:

n = \frac{7}{5}, n = \frac{4}{3}

3 s^{2} + 6 s + 2 = 0

4 5^{2} + 6 0^{2} = x^{2}

- \frac{2}{3} (x - 1) (x + 3) = 0

- y^{2} - 4 y - 4 = 0

12 x - x^{2} = 36

What are the solutions to the equation 120n^2-328n=-224 ?
The solutions to the equation 120n^2-328n=-224 are n= 7/5 ,n= 4/3
Find the zeros of 120n^2-328n=-224
The zeros of 120n^2-328n=-224 are n= 7/5 ,n= 4/3