What are the solutions to the equation x^2-2x=-10 ?

The solutions to the equation x^2-2x=-10 are x=1+3i,x=1-3i

Find the zeros of x^2-2x=-10

The zeros of x^2-2x=-10 are x=1+3i,x=1-3i

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x^{2} - 2 x = - 10

x = 1 + 3 i, x = 1 - 3 i

Solution steps

x^{2} - 2 x = - 10

Move

10

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 10 : 6 i

x_{1, 2} = \frac{- ( - 2 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6 i}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 6 i}{2 \cdot 1}

x = \frac{- ( - 2 ) + 6 i}{2 \cdot 1} : 1 + 3 i

x = \frac{- ( - 2 ) - 6 i}{2 \cdot 1} : 1 - 3 i

The solutions to the quadratic equation are:

x = 1 + 3 i, x = 1 - 3 i

12 - \frac{5 x}{2} \geq \frac{x}{4} + 1

49 a^{2} b^{2} - 14 ab + 1

- t^{5} + t^{3}

\frac{0}{7}

x \to \infty lim (e^{x})

What are the solutions to the equation x^2-2x=-10 ?
The solutions to the equation x^2-2x=-10 are x=1+3i,x=1-3i
Find the zeros of x^2-2x=-10
The zeros of x^2-2x=-10 are x=1+3i,x=1-3i