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

The solutions to the equation 12x^2+16x-3=0 are x= 1/6 ,x=-3/2

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

The zeros of 12x^2+16x-3=0 are x= 1/6 ,x=-3/2

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

x = \frac{1}{6}, x = - \frac{3}{2}

Decimal Notation

x = 0.16666 \dots, x = - 1.5

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 12 ( - 3 )}{2 \cdot 12}

1 6^{2} - 4 \cdot 12 (- 3) = 20

x_{1, 2} = \frac{- 16 \pm 20}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 16 + 20}{2 \cdot 12}, x_{2} = \frac{- 16 - 20}{2 \cdot 12}

x = \frac{- 16 + 20}{2 \cdot 12} : \frac{1}{6}

x = \frac{- 16 - 20}{2 \cdot 12} : - \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{6}, x = - \frac{3}{2}

x^{2} - 4 x = - 29

121 x^{2} - 8 = 0

- x^{2} + 4 x + 2

lo g_{3} (2) x^{2} = lo g_{3} (72)

2 x^{2} + 24 x - 60 = x (x + 13)

What are the solutions to the equation 12x^2+16x-3=0 ?
The solutions to the equation 12x^2+16x-3=0 are x= 1/6 ,x=-3/2
Find the zeros of 12x^2+16x-3=0
The zeros of 12x^2+16x-3=0 are x= 1/6 ,x=-3/2