What are the solutions to the equation 5n^2+7n-3=4n ?

The solutions to the equation 5n^2+7n-3=4n are n=(-3+sqrt(69))/(10),n=(-3-sqrt(69))/(10)

Find the zeros of 5n^2+7n-3=4n

The zeros of 5n^2+7n-3=4n are n=(-3+sqrt(69))/(10),n=(-3-sqrt(69))/(10)

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5 n^{2} + 7 n - 3 = 4 n

n = \frac{- 3 + 69}{10}, n = \frac{- 3 - 69}{10}

Solution steps

5 n^{2} + 7 n - 3 = 4 n

Move

4 n

to the left side

5 n^{2} + 3 n - 3 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 3) = 69

n_{1, 2} = \frac{- 3 \pm 69}{2 \cdot 5}

Separate the solutions

n_{1} = \frac{- 3 + 69}{2 \cdot 5}, n_{2} = \frac{- 3 - 69}{2 \cdot 5}

n = \frac{- 3 + 69}{2 \cdot 5} : \frac{- 3 + 69}{10}

n = \frac{- 3 - 69}{2 \cdot 5} : \frac{- 3 - 69}{10}

The solutions to the quadratic equation are:

n = \frac{- 3 + 69}{10}, n = \frac{- 3 - 69}{10}

x^{2} - 11 x + 41 = 19

4 x^{2} - 3 (5 x - 1) = 0

- 0.1 x^{2} + 14 x + 3 = 1000

x (x - 5) = 4 x

16 x^{2} + 22 = 0

What are the solutions to the equation 5n^2+7n-3=4n ?
The solutions to the equation 5n^2+7n-3=4n are n=(-3+sqrt(69))/(10),n=(-3-sqrt(69))/(10)
Find the zeros of 5n^2+7n-3=4n
The zeros of 5n^2+7n-3=4n are n=(-3+sqrt(69))/(10),n=(-3-sqrt(69))/(10)