What are the solutions to the equation 4x^2+8x-12=0 ?

The solutions to the equation 4x^2+8x-12=0 are x=1,x=-3

Find the zeros of 4x^2+8x-12=0

The zeros of 4x^2+8x-12=0 are x=1,x=-3

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4 x^{2} + 8 x - 12 = 0

x = 1, x = - 3

Solution steps

4 x^{2} + 8 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 ( - 12 )}{2 \cdot 4}

8^{2} - 4 \cdot 4 (- 12) = 16

x_{1, 2} = \frac{- 8 \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 8 + 16}{2 \cdot 4}, x_{2} = \frac{- 8 - 16}{2 \cdot 4}

x = \frac{- 8 + 16}{2 \cdot 4} : 1

x = \frac{- 8 - 16}{2 \cdot 4} : - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

x^{2} y + x + x y^{2} + y + 2 x y + 2

ab - b^{2} + 3 a - 3 b

- 3 z \leq - 9

\frac{( 2 x + 1 )}{5} = \frac{( x + 1 )}{5}

What are the solutions to the equation 4x^2+8x-12=0 ?
The solutions to the equation 4x^2+8x-12=0 are x=1,x=-3
Find the zeros of 4x^2+8x-12=0
The zeros of 4x^2+8x-12=0 are x=1,x=-3