What are the solutions to the equation 48-(10-3+x^2)+2xx=4 ?

The solutions to the equation 48-(10-3+x^2)+2xx=4 are x=sqrt(37)i,x=-sqrt(37)i

Find the zeros of 48-(10-3+x^2)+2xx=4

The zeros of 48-(10-3+x^2)+2xx=4 are x=sqrt(37)i,x=-sqrt(37)i

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48 - (10 - 3 + x^{2}) + 2 xx = 4

x = 37 i, x = - 37 i

Solution steps

48 - (10 - 3 + x^{2}) + 2 xx = 4

Expand

48 - (10 - 3 + x^{2}) + 2 xx : x^{2} + 41

x^{2} + 41 = 4

Move

4

to the left side

x^{2} + 37 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot 37}{2 \cdot 1}

Simplify

0^{2} - 4 \cdot 1 \cdot 37 : 237 i

x_{1, 2} = \frac{- 0 \pm 2 37 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 2 37 i}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 37 i}{2 \cdot 1}

x = \frac{- 0 + 2 37 i}{2 \cdot 1} : 37 i

x = \frac{- 0 - 2 37 i}{2 \cdot 1} : - 37 i

The solutions to the quadratic equation are:

x = 37 i, x = - 37 i

- x^{2} - 3 x = 4

3 m^{2} - 2 = 1

x^{2} + 14 x - 3 = 0

x (x - 4) + (m + 1) = 0

(y - 1) (y - 2) = \frac{1}{2} y^{2}

What are the solutions to the equation 48-(10-3+x^2)+2xx=4 ?
The solutions to the equation 48-(10-3+x^2)+2xx=4 are x=sqrt(37)i,x=-sqrt(37)i
Find the zeros of 48-(10-3+x^2)+2xx=4
The zeros of 48-(10-3+x^2)+2xx=4 are x=sqrt(37)i,x=-sqrt(37)i