What are the solutions to the equation (x-4)^2+(2x-15+2)^2=5 ?

The solutions to the equation (x-4)^2+(2x-15+2)^2=5 are x=6

Find the zeros of (x-4)^2+(2x-15+2)^2=5

The zeros of (x-4)^2+(2x-15+2)^2=5 are x=6

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(x - 4)^{2} + (2 x - 15 + 2)^{2} = 5

x = 6

Solution steps

(x - 4)^{2} + (2 x - 15 + 2)^{2} = 5

Expand

(x - 4)^{2} + (2 x - 15 + 2)^{2} : 5 x^{2} - 60 x + 185

5 x^{2} - 60 x + 185 = 5

Move

5

to the left side

5 x^{2} - 60 x + 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 60 ) \pm ( - 60 ) ^{2} - 4 \cdot 5 \cdot 180}{2 \cdot 5}

(- 60)^{2} - 4 \cdot 5 \cdot 180 = 0

x_{1, 2} = \frac{- ( - 60 ) \pm 0}{2 \cdot 5}

x = \frac{- ( - 60 )}{2 \cdot 5}

\frac{- ( - 60 )}{2 \cdot 5} = 6

x = 6

The solution to the quadratic equation is:

x = 6

- 4.91 x^{2} + 2 = 0

x^{2} + (x + 1)^{2} + (x + 2)^{2} = 110

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

70840 - 188 d - 2.35 d^{2} = 1228.8 d + 30.72 d^{2}

u^{2} - 16 u

What are the solutions to the equation (x-4)^2+(2x-15+2)^2=5 ?
The solutions to the equation (x-4)^2+(2x-15+2)^2=5 are x=6
Find the zeros of (x-4)^2+(2x-15+2)^2=5
The zeros of (x-4)^2+(2x-15+2)^2=5 are x=6