What are the solutions to the equation 2q^2+q-1596=0 ?

The solutions to the equation 2q^2+q-1596=0 are q=28,q=-57/2

Find the zeros of 2q^2+q-1596=0

The zeros of 2q^2+q-1596=0 are q=28,q=-57/2

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2 q^{2} + q - 1596 = 0

q = 28, q = - \frac{57}{2}

Decimal Notation

q = 28, q = - 28.5

Solution steps

2 q^{2} + q - 1596 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1596 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1596) = 113

q_{1, 2} = \frac{- 1 \pm 113}{2 \cdot 2}

Separate the solutions

q_{1} = \frac{- 1 + 113}{2 \cdot 2}, q_{2} = \frac{- 1 - 113}{2 \cdot 2}

q = \frac{- 1 + 113}{2 \cdot 2} : 28

q = \frac{- 1 - 113}{2 \cdot 2} : - \frac{57}{2}

The solutions to the quadratic equation are:

q = 28, q = - \frac{57}{2}

x, T (x) = x^{2} + 30 x + 200

x^{2} + 2 x + (- 15) = 0

2 x^{2} + x = 5

5 x^{2} + 400 = 0

(4 x + 9)^{2} = 36

What are the solutions to the equation 2q^2+q-1596=0 ?
The solutions to the equation 2q^2+q-1596=0 are q=28,q=-57/2
Find the zeros of 2q^2+q-1596=0
The zeros of 2q^2+q-1596=0 are q=28,q=-57/2