What are the solutions to the equation (x+8)^2+(x-10)^2=-156 ?

The solutions to the equation (x+8)^2+(x-10)^2=-156 are x=1+sqrt(159)i,x=1-sqrt(159)i

Find the zeros of (x+8)^2+(x-10)^2=-156

The zeros of (x+8)^2+(x-10)^2=-156 are x=1+sqrt(159)i,x=1-sqrt(159)i

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(x + 8)^{2} + (x - 10)^{2} = - 156

x = 1 + 159 i, x = 1 - 159 i

Solution steps

(x + 8)^{2} + (x - 10)^{2} = - 156

Expand

(x + 8)^{2} + (x - 10)^{2} : 2 x^{2} - 4 x + 164

2 x^{2} - 4 x + 164 = - 156

Move

156

to the left side

2 x^{2} - 4 x + 320 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 320}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 \cdot 320 : 4159 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 159 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 159 i}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 4 159 i}{2 \cdot 2}

x = \frac{- ( - 4 ) + 4 159 i}{2 \cdot 2} : 1 + 159 i

x = \frac{- ( - 4 ) - 4 159 i}{2 \cdot 2} : 1 - 159 i

The solutions to the quadratic equation are:

x = 1 + 159 i, x = 1 - 159 i

x^{2} + 16 x + 47 = 0

b^{2} - 8 b + 15 = 0

t^{2} - 2 t + 1 = 36

x, x^{2} + y^{2} + 6 x - 4 y + 13 = 0

3 t^{2} - t - 6

What are the solutions to the equation (x+8)^2+(x-10)^2=-156 ?
The solutions to the equation (x+8)^2+(x-10)^2=-156 are x=1+sqrt(159)i,x=1-sqrt(159)i
Find the zeros of (x+8)^2+(x-10)^2=-156
The zeros of (x+8)^2+(x-10)^2=-156 are x=1+sqrt(159)i,x=1-sqrt(159)i