What are the solutions to the equation 3q^2+128q-600=0 ?

The solutions to the equation 3q^2+128q-600=0 are q=(2(sqrt(1474)-32))/3 ,q=-(2(32+sqrt(1474)))/3

Find the zeros of 3q^2+128q-600=0

The zeros of 3q^2+128q-600=0 are q=(2(sqrt(1474)-32))/3 ,q=-(2(32+sqrt(1474)))/3

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3 q^{2} + 128 q - 600 = 0

q = \frac{2 ( 1474 - 32 )}{3}, q = - \frac{2 ( 32 + 1474 )}{3}

Decimal Notation

q = 4.26180 \dots, q = - 46.92847 \dots

Solution steps

3 q^{2} + 128 q - 600 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 \cdot 3 ( - 600 )}{2 \cdot 3}

12 8^{2} - 4 \cdot 3 (- 600) = 41474

q_{1, 2} = \frac{- 128 \pm 4 1474}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- 128 + 4 1474}{2 \cdot 3}, q_{2} = \frac{- 128 - 4 1474}{2 \cdot 3}

q = \frac{- 128 + 4 1474}{2 \cdot 3} : \frac{2 ( 1474 - 32 )}{3}

q = \frac{- 128 - 4 1474}{2 \cdot 3} : - \frac{2 ( 32 + 1474 )}{3}

The solutions to the quadratic equation are:

q = \frac{2 ( 1474 - 32 )}{3}, q = - \frac{2 ( 32 + 1474 )}{3}

x^{2} - 5 x - 40 = 0

x^{2} + 5 x + 11 = 13 x

2 W^{2} - 3 W - 27 = 0

r^{2} - r - 11 = 0

- 3 x + 2 x^{2} = 2

What are the solutions to the equation 3q^2+128q-600=0 ?
The solutions to the equation 3q^2+128q-600=0 are q=(2(sqrt(1474)-32))/3 ,q=-(2(32+sqrt(1474)))/3
Find the zeros of 3q^2+128q-600=0
The zeros of 3q^2+128q-600=0 are q=(2(sqrt(1474)-32))/3 ,q=-(2(32+sqrt(1474)))/3