What are the solutions to the equation x^2+16x=-39 ?

The solutions to the equation x^2+16x=-39 are x=-3,x=-13

Find the zeros of x^2+16x=-39

The zeros of x^2+16x=-39 are x=-3,x=-13

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x^{2} + 16 x = - 39

x = - 3, x = - 13

Solution steps

x^{2} + 16 x = - 39

Move

39

to the left side

x^{2} + 16 x + 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 39}{2 \cdot 1}

1 6^{2} - 4 \cdot 1 \cdot 39 = 10

x_{1, 2} = \frac{- 16 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 10}{2 \cdot 1}, x_{2} = \frac{- 16 - 10}{2 \cdot 1}

x = \frac{- 16 + 10}{2 \cdot 1} : - 3

x = \frac{- 16 - 10}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

x = - 3, x = - 13

7 x^{2} - 41 x + 20 = - 2 x

\frac{- 700}{- 100}

4 x + 3 > 3 x + 7

3 t - 1 < 10 or 2 t + 4

2^{3} \times (- 3)^{3}

What are the solutions to the equation x^2+16x=-39 ?
The solutions to the equation x^2+16x=-39 are x=-3,x=-13
Find the zeros of x^2+16x=-39
The zeros of x^2+16x=-39 are x=-3,x=-13