What are the solutions to the equation x^2-7x+25=0 ?

The solutions to the equation x^2-7x+25=0 are x= 7/2+i(sqrt(51))/2 ,x= 7/2-i(sqrt(51))/2

Find the zeros of x^2-7x+25=0

The zeros of x^2-7x+25=0 are x= 7/2+i(sqrt(51))/2 ,x= 7/2-i(sqrt(51))/2

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x^{2} - 7 x + 25 = 0

x = \frac{7}{2} + i \frac{51}{2}, x = \frac{7}{2} - i \frac{51}{2}

Solution steps

x^{2} - 7 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

Simplify

(- 7)^{2} - 4 \cdot 1 \cdot 25 : 51 i

x_{1, 2} = \frac{- ( - 7 ) \pm 51 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 51 i}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 51 i}{2 \cdot 1}

x = \frac{- ( - 7 ) + 51 i}{2 \cdot 1} : \frac{7}{2} + i \frac{51}{2}

x = \frac{- ( - 7 ) - 51 i}{2 \cdot 1} : \frac{7}{2} - i \frac{51}{2}

The solutions to the quadratic equation are:

x = \frac{7}{2} + i \frac{51}{2}, x = \frac{7}{2} - i \frac{51}{2}

ln (3 x) \leq 5

y^{4} - 3 y^{2} - 28

- 699 [- 2 (132 + 84) - 4 (- 52 + (- 108))] + 499

\frac{3 + 9 i}{1 + i}

x^{2} - 11 x + 68 = 5 x + 5

What are the solutions to the equation x^2-7x+25=0 ?
The solutions to the equation x^2-7x+25=0 are x= 7/2+i(sqrt(51))/2 ,x= 7/2-i(sqrt(51))/2
Find the zeros of x^2-7x+25=0
The zeros of x^2-7x+25=0 are x= 7/2+i(sqrt(51))/2 ,x= 7/2-i(sqrt(51))/2