What are the solutions to the equation 12x^2+36x-48=0 ?

The solutions to the equation 12x^2+36x-48=0 are x=1,x=-4

Find the zeros of 12x^2+36x-48=0

The zeros of 12x^2+36x-48=0 are x=1,x=-4

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12 x^{2} + 36 x - 48 = 0

x = 1, x = - 4

Solution steps

12 x^{2} + 36 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 12 ( - 48 )}{2 \cdot 12}

3 6^{2} - 4 \cdot 12 (- 48) = 60

x_{1, 2} = \frac{- 36 \pm 60}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 36 + 60}{2 \cdot 12}, x_{2} = \frac{- 36 - 60}{2 \cdot 12}

x = \frac{- 36 + 60}{2 \cdot 12} : 1

x = \frac{- 36 - 60}{2 \cdot 12} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

6 x (x - 3) = 0

4 x^{2} - 6 x + \frac{9}{4} = 0

x, y = x^{2} - 2 x + 1

2 x^{2} + 4 = 2 (x + 2) - 3 x^{2}

\frac{( y + 1 ) ^{2}}{49} = 1

What are the solutions to the equation 12x^2+36x-48=0 ?
The solutions to the equation 12x^2+36x-48=0 are x=1,x=-4
Find the zeros of 12x^2+36x-48=0
The zeros of 12x^2+36x-48=0 are x=1,x=-4