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

The solutions to the equation 2x^2+16x-12=0 are x=-4+sqrt(22),x=-4-sqrt(22)

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

The zeros of 2x^2+16x-12=0 are x=-4+sqrt(22),x=-4-sqrt(22)

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

x = - 4 + 22, x = - 4 - 22

Decimal Notation

x = 0.69041 \dots, x = - 8.69041 \dots

Solution steps

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

Solve with the quadratic formula

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

1 6^{2} - 4 \cdot 2 (- 12) = 422

x_{1, 2} = \frac{- 16 \pm 4 22}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 4 22}{2 \cdot 2}, x_{2} = \frac{- 16 - 4 22}{2 \cdot 2}

x = \frac{- 16 + 4 22}{2 \cdot 2} : - 4 + 22

x = \frac{- 16 - 4 22}{2 \cdot 2} : - 4 - 22

The solutions to the quadratic equation are:

x = - 4 + 22, x = - 4 - 22

- 17 x^{2} + 14 + 6 x = 6 x - 7 x^{2}

y, z = (1 - 2 y) (3 y + 2)

4 x^{2} + 5 x + 10 = 0

(x + 5)^{2} = 11

r^{2} + 3 r - 4 = 0

What are the solutions to the equation 2x^2+16x-12=0 ?
The solutions to the equation 2x^2+16x-12=0 are x=-4+sqrt(22),x=-4-sqrt(22)
Find the zeros of 2x^2+16x-12=0
The zeros of 2x^2+16x-12=0 are x=-4+sqrt(22),x=-4-sqrt(22)