What are the solutions to the equation 5x^2-2x+13=9 ?

The solutions to the equation 5x^2-2x+13=9 are x= 1/5+i(sqrt(19))/5 ,x= 1/5-i(sqrt(19))/5

Find the zeros of 5x^2-2x+13=9

The zeros of 5x^2-2x+13=9 are x= 1/5+i(sqrt(19))/5 ,x= 1/5-i(sqrt(19))/5

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5 x^{2} - 2 x + 13 = 9

x = \frac{1}{5} + i \frac{19}{5}, x = \frac{1}{5} - i \frac{19}{5}

Solution steps

5 x^{2} - 2 x + 13 = 9

Move

9

to the left side

5 x^{2} - 2 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

Simplify

(- 2)^{2} - 4 \cdot 5 \cdot 4 : 219 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 19 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 19 i}{2 \cdot 5}, x_{2} = \frac{- ( - 2 ) - 2 19 i}{2 \cdot 5}

x = \frac{- ( - 2 ) + 2 19 i}{2 \cdot 5} : \frac{1}{5} + i \frac{19}{5}

x = \frac{- ( - 2 ) - 2 19 i}{2 \cdot 5} : \frac{1}{5} - i \frac{19}{5}

The solutions to the quadratic equation are:

x = \frac{1}{5} + i \frac{19}{5}, x = \frac{1}{5} - i \frac{19}{5}

(x + 2) (x - 9) = 0

(x + 1)^{2} + 14 = 15

4 (x - 8)^{2} + 11 = 15

x, z = x^{2} - 3 y^{2} + 5 x y

4 x^{2} - 18 x = 0

What are the solutions to the equation 5x^2-2x+13=9 ?
The solutions to the equation 5x^2-2x+13=9 are x= 1/5+i(sqrt(19))/5 ,x= 1/5-i(sqrt(19))/5
Find the zeros of 5x^2-2x+13=9
The zeros of 5x^2-2x+13=9 are x= 1/5+i(sqrt(19))/5 ,x= 1/5-i(sqrt(19))/5