What are the solutions to the equation x^2+14x+48=0 ?

The solutions to the equation x^2+14x+48=0 are x=-6,x=-8

Find the zeros of x^2+14x+48=0

The zeros of x^2+14x+48=0 are x=-6,x=-8

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x^{2} + 14 x + 48 = 0

x = - 6, x = - 8

Solution steps

x^{2} + 14 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 48 = 2

x_{1, 2} = \frac{- 14 \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 14 + 2}{2 \cdot 1}, x_{2} = \frac{- 14 - 2}{2 \cdot 1}

x = \frac{- 14 + 2}{2 \cdot 1} : - 6

x = \frac{- 14 - 2}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 6, x = - 8

9 x^{2} - 1 = 24

(2 a - 3 b) 4 b - 3 a (a - 2 b)

3 a^{2} - 4 ab - 20 b^{2}

4 c^{2} - 9 d^{2}

\frac{r}{2} > 2

What are the solutions to the equation x^2+14x+48=0 ?
The solutions to the equation x^2+14x+48=0 are x=-6,x=-8
Find the zeros of x^2+14x+48=0
The zeros of x^2+14x+48=0 are x=-6,x=-8