What are the solutions to the equation x^2-x+42=0 ?

The solutions to the equation x^2-x+42=0 are x= 1/2+i(sqrt(167))/2 ,x= 1/2-i(sqrt(167))/2

Find the zeros of x^2-x+42=0

The zeros of x^2-x+42=0 are x= 1/2+i(sqrt(167))/2 ,x= 1/2-i(sqrt(167))/2

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x^{2} - x + 42 = 0

x = \frac{1}{2} + i \frac{167}{2}, x = \frac{1}{2} - i \frac{167}{2}

Solution steps

x^{2} - x + 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 42 : 167 i

x_{1, 2} = \frac{- ( - 1 ) \pm 167 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 167 i}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 167 i}{2 \cdot 1}

x = \frac{- ( - 1 ) + 167 i}{2 \cdot 1} : \frac{1}{2} + i \frac{167}{2}

x = \frac{- ( - 1 ) - 167 i}{2 \cdot 1} : \frac{1}{2} - i \frac{167}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{167}{2}, x = \frac{1}{2} - i \frac{167}{2}

3 x^{2} + 15 x = 0

2 x^{2} + 5 x + 6 = - 2 x

x, x^{2} + y^{2} = s^{2}

5 x^{2} - 25 x + 1 = 0

x^{2} - 11 x + 18 = 0

What are the solutions to the equation x^2-x+42=0 ?
The solutions to the equation x^2-x+42=0 are x= 1/2+i(sqrt(167))/2 ,x= 1/2-i(sqrt(167))/2
Find the zeros of x^2-x+42=0
The zeros of x^2-x+42=0 are x= 1/2+i(sqrt(167))/2 ,x= 1/2-i(sqrt(167))/2