What are the solutions to the equation 25=x^2+14x+49 ?

The solutions to the equation 25=x^2+14x+49 are x=-2,x=-12

Find the zeros of 25=x^2+14x+49

The zeros of 25=x^2+14x+49 are x=-2,x=-12

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25 = x^{2} + 14 x + 49

x = - 2, x = - 12

Solution steps

25 = x^{2} + 14 x + 49

Switch sides

x^{2} + 14 x + 49 = 25

Move

25

to the left side

x^{2} + 14 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 24 = 10

x_{1, 2} = \frac{- 14 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 14 + 10}{2 \cdot 1}, x_{2} = \frac{- 14 - 10}{2 \cdot 1}

x = \frac{- 14 + 10}{2 \cdot 1} : - 2

x = \frac{- 14 - 10}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = - 2, x = - 12

2 x^{2} - 3 x = - 2

115 + 12.25 t - 4.9 t^{2}

(m - 3) x^{2} + (m - 7) x + 2 = 0

2 x^{2} - 3 x = - 5

5 x^{2} + 7 = 27

What are the solutions to the equation 25=x^2+14x+49 ?
The solutions to the equation 25=x^2+14x+49 are x=-2,x=-12
Find the zeros of 25=x^2+14x+49
The zeros of 25=x^2+14x+49 are x=-2,x=-12