What are the solutions to the equation 4n^2+13n=-3 ?

The solutions to the equation 4n^2+13n=-3 are n=-1/4 ,n=-3

Find the zeros of 4n^2+13n=-3

The zeros of 4n^2+13n=-3 are n=-1/4 ,n=-3

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4 n^{2} + 13 n = - 3

n = - \frac{1}{4}, n = - 3

Decimal Notation

n = - 0.25, n = - 3

Solution steps

4 n^{2} + 13 n = - 3

Move

3

to the left side

4 n^{2} + 13 n + 3 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

1 3^{2} - 4 \cdot 4 \cdot 3 = 11

n_{1, 2} = \frac{- 13 \pm 11}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- 13 + 11}{2 \cdot 4}, n_{2} = \frac{- 13 - 11}{2 \cdot 4}

n = \frac{- 13 + 11}{2 \cdot 4} : - \frac{1}{4}

n = \frac{- 13 - 11}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

n = - \frac{1}{4}, n = - 3

7 x - 8 = - 3 x^{2}

15 x^{2} - x - 6 = 0

- 3 (3 x + 15) - (10 + x) = 35

x^{2} - 1 = x^{2} - 1

7 t - 6 v^{2} + t v - 42 v

What are the solutions to the equation 4n^2+13n=-3 ?
The solutions to the equation 4n^2+13n=-3 are n=-1/4 ,n=-3
Find the zeros of 4n^2+13n=-3
The zeros of 4n^2+13n=-3 are n=-1/4 ,n=-3