What are the solutions to the equation x^2+10x+25=121 ?

The solutions to the equation x^2+10x+25=121 are x=6,x=-16

Find the zeros of x^2+10x+25=121

The zeros of x^2+10x+25=121 are x=6,x=-16

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x^{2} + 10 x + 25 = 121

x = 6, x = - 16

Solution steps

x^{2} + 10 x + 25 = 121

Move

121

to the left side

x^{2} + 10 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 96 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 96) = 22

x_{1, 2} = \frac{- 10 \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 22}{2 \cdot 1}, x_{2} = \frac{- 10 - 22}{2 \cdot 1}

x = \frac{- 10 + 22}{2 \cdot 1} : 6

x = \frac{- 10 - 22}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

x = 6, x = - 16

x^{2} - 4 x + 13

x^{2} + x - 4 b x = b - 3 b^{2}

x^{2} - 6 x + 10 = 0

- y^{2} + 2 = 12 y - 7

9 x^{2} + 6 x + 1 = 4

What are the solutions to the equation x^2+10x+25=121 ?
The solutions to the equation x^2+10x+25=121 are x=6,x=-16
Find the zeros of x^2+10x+25=121
The zeros of x^2+10x+25=121 are x=6,x=-16