What are the solutions to the equation x^2+7x+12=0 ?

The solutions to the equation x^2+7x+12=0 are x=-3,x=-4

Find the zeros of x^2+7x+12=0

The zeros of x^2+7x+12=0 are x=-3,x=-4

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x^{2} + 7 x + 12 = 0

x = - 3, x = - 4

Solution steps

x^{2} + 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

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

Separate the solutions

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

x = \frac{- 7 + 1}{2 \cdot 1} : - 3

x = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 3, x = - 4

\frac{2}{3} (6 - 9 x) = 16

8 x^{3} - 64 x^{2} + x - 8

\frac{d y}{d t} - y = - \frac{t}{y}

\frac{( 9 s t ) ^{\frac{3}{2}}}{( 27 s ^{3} t ^{- 4} ) ^{\frac{2}{3}}}

x - 12 + 4 x = - 2 x - 28 + x

What are the solutions to the equation x^2+7x+12=0 ?
The solutions to the equation x^2+7x+12=0 are x=-3,x=-4
Find the zeros of x^2+7x+12=0
The zeros of x^2+7x+12=0 are x=-3,x=-4