What are the solutions to the equation 224-24q+1.5q^2=0 ?

The solutions to the equation 224-24q+1.5q^2=0 are q=8+i(16sqrt(3))/3 ,q=8-i(16sqrt(3))/3

Find the zeros of 224-24q+1.5q^2=0

The zeros of 224-24q+1.5q^2=0 are q=8+i(16sqrt(3))/3 ,q=8-i(16sqrt(3))/3

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224 - 24 q + 1.5 q^{2} = 0

q = 8 + i \frac{16 3}{3}, q = 8 - i \frac{16 3}{3}

Solution steps

224 - 24 q + 1.5 q^{2} = 0

Multiply both sides by

10

2240 - 240 q + 15 q^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

15 q^{2} - 240 q + 2240 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 240 ) \pm ( - 240 ) ^{2} - 4 \cdot 15 \cdot 2240}{2 \cdot 15}

Simplify

(- 240)^{2} - 4 \cdot 15 \cdot 2240 : 1603 i

q_{1, 2} = \frac{- ( - 240 ) \pm 160 3 i}{2 \cdot 15}

Separate the solutions

q_{1} = \frac{- ( - 240 ) + 160 3 i}{2 \cdot 15}, q_{2} = \frac{- ( - 240 ) - 160 3 i}{2 \cdot 15}

q = \frac{- ( - 240 ) + 160 3 i}{2 \cdot 15} : 8 + i \frac{16 3}{3}

q = \frac{- ( - 240 ) - 160 3 i}{2 \cdot 15} : 8 - i \frac{16 3}{3}

The solutions to the quadratic equation are:

q = 8 + i \frac{16 3}{3}, q = 8 - i \frac{16 3}{3}

4 + 20 x^{2} + 8 x = 0

- 4 x^{2} - 7 x + 9

2 x^{2} + 8 x

x^{2} - x - 1 = 0

y^{2} + 14 y + 48 = 0

What are the solutions to the equation 224-24q+1.5q^2=0 ?
The solutions to the equation 224-24q+1.5q^2=0 are q=8+i(16sqrt(3))/3 ,q=8-i(16sqrt(3))/3
Find the zeros of 224-24q+1.5q^2=0
The zeros of 224-24q+1.5q^2=0 are q=8+i(16sqrt(3))/3 ,q=8-i(16sqrt(3))/3