What are the solutions to the equation 8x^2+64x+120=0 ?

The solutions to the equation 8x^2+64x+120=0 are x=-3,x=-5

Find the zeros of 8x^2+64x+120=0

The zeros of 8x^2+64x+120=0 are x=-3,x=-5

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8 x^{2} + 64 x + 120 = 0

x = - 3, x = - 5

Solution steps

8 x^{2} + 64 x + 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 64 \pm 6 4 ^{2} - 4 \cdot 8 \cdot 120}{2 \cdot 8}

6 4^{2} - 4 \cdot 8 \cdot 120 = 16

x_{1, 2} = \frac{- 64 \pm 16}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 64 + 16}{2 \cdot 8}, x_{2} = \frac{- 64 - 16}{2 \cdot 8}

x = \frac{- 64 + 16}{2 \cdot 8} : - 3

x = \frac{- 64 - 16}{2 \cdot 8} : - 5

The solutions to the quadratic equation are:

x = - 3, x = - 5

(\frac{4 x y}{5 x ^{2} y ^{- 2}})^{2}

36 - 4.9 (x)^{2} = 10

(\frac{2}{7})^{2}

- 6 \leq \frac{2 x + 11}{7} < 2

(\frac{x ^{2}}{y ^{3}})^{5}

What are the solutions to the equation 8x^2+64x+120=0 ?
The solutions to the equation 8x^2+64x+120=0 are x=-3,x=-5
Find the zeros of 8x^2+64x+120=0
The zeros of 8x^2+64x+120=0 are x=-3,x=-5