What are the solutions to the equation x^2-3x+7=0 ?

The solutions to the equation x^2-3x+7=0 are x= 3/2+i(sqrt(19))/2 ,x= 3/2-i(sqrt(19))/2

Find the zeros of x^2-3x+7=0

The zeros of x^2-3x+7=0 are x= 3/2+i(sqrt(19))/2 ,x= 3/2-i(sqrt(19))/2

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x^{2} - 3 x + 7 = 0

x = \frac{3}{2} + i \frac{19}{2}, x = \frac{3}{2} - i \frac{19}{2}

Solution steps

x^{2} - 3 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

Simplify

(- 3)^{2} - 4 \cdot 1 \cdot 7 : 19 i

x_{1, 2} = \frac{- ( - 3 ) \pm 19 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 19 i}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 19 i}{2 \cdot 1}

x = \frac{- ( - 3 ) + 19 i}{2 \cdot 1} : \frac{3}{2} + i \frac{19}{2}

x = \frac{- ( - 3 ) - 19 i}{2 \cdot 1} : \frac{3}{2} - i \frac{19}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{19}{2}, x = \frac{3}{2} - i \frac{19}{2}

2 (2 x - 4) + 3 (3 x + 6) = 49

\frac{3}{4} - \frac{5}{3}

x^{2} + 20 x - 40 > 0

9 u^{2} + 48 u + 64

\frac{1}{- 2 i}

What are the solutions to the equation x^2-3x+7=0 ?
The solutions to the equation x^2-3x+7=0 are x= 3/2+i(sqrt(19))/2 ,x= 3/2-i(sqrt(19))/2
Find the zeros of x^2-3x+7=0
The zeros of x^2-3x+7=0 are x= 3/2+i(sqrt(19))/2 ,x= 3/2-i(sqrt(19))/2