What are the solutions to the equation 2x^2+16x-24=0 ?

The solutions to the equation 2x^2+16x-24=0 are x=2(sqrt(7)-2),x=-2(2+sqrt(7))

Find the zeros of 2x^2+16x-24=0

The zeros of 2x^2+16x-24=0 are x=2(sqrt(7)-2),x=-2(2+sqrt(7))

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2 x^{2} + 16 x - 24 = 0

x = 2 (7 - 2), x = - 2 (2 + 7)

Decimal Notation

x = 1.29150 \dots, x = - 9.29150 \dots

Solution steps

2 x^{2} + 16 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 (- 24) = 87

x_{1, 2} = \frac{- 16 \pm 8 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 16 + 8 7}{2 \cdot 2}, x_{2} = \frac{- 16 - 8 7}{2 \cdot 2}

x = \frac{- 16 + 8 7}{2 \cdot 2} : 2 (7 - 2)

x = \frac{- 16 - 8 7}{2 \cdot 2} : - 2 (2 + 7)

The solutions to the quadratic equation are:

x = 2 (7 - 2), x = - 2 (2 + 7)

t, f n (t) = 5 t - t^{2}

x^{2} + 18 x = 6 x - 35

(x + 3)^{2} = (x - 3)^{2} + (x - 6) (x + 6) + 63

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

16 p^{2} - 24 p + 5 = 0

What are the solutions to the equation 2x^2+16x-24=0 ?
The solutions to the equation 2x^2+16x-24=0 are x=2(sqrt(7)-2),x=-2(2+sqrt(7))
Find the zeros of 2x^2+16x-24=0
The zeros of 2x^2+16x-24=0 are x=2(sqrt(7)-2),x=-2(2+sqrt(7))