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

(a + b)^{2} + (a - b)^{2}

- \frac{2}{4}

\frac{11 !}{11 ! ( 11 - 11 ) !}

m + \frac{2}{3} = \frac{1}{4} m - 1

\frac{1}{2} (1 - 0.95)

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