What are the solutions to the equation -16x+3x^2-12=0 ?

The solutions to the equation -16x+3x^2-12=0 are x=6,x=-2/3

Find the zeros of -16x+3x^2-12=0

The zeros of -16x+3x^2-12=0 are x=6,x=-2/3

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- 16 x + 3 x^{2} - 12 = 0

x = 6, x = - \frac{2}{3}

Solution steps

- 16 x + 3 x^{2} - 12 = 0

Write in the standard form

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

3 x^{2} - 16 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

(- 16)^{2} - 4 \cdot 3 (- 12) = 20

x_{1, 2} = \frac{- ( - 16 ) \pm 20}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 20}{2 \cdot 3}, x_{2} = \frac{- ( - 16 ) - 20}{2 \cdot 3}

x = \frac{- ( - 16 ) + 20}{2 \cdot 3} : 6

x = \frac{- ( - 16 ) - 20}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = 6, x = - \frac{2}{3}

4 x - 2 < 10

1 \leq 2

x^{2} + 12 x = - 36

(x + 11) (x + 5) = - 4

ln (x - 11) < 0

What are the solutions to the equation -16x+3x^2-12=0 ?
The solutions to the equation -16x+3x^2-12=0 are x=6,x=-2/3
Find the zeros of -16x+3x^2-12=0
The zeros of -16x+3x^2-12=0 are x=6,x=-2/3