What are the solutions to the equation 5x^2-11x+8=0 ?

The solutions to the equation 5x^2-11x+8=0 are x= 11/10+i(sqrt(39))/(10),x= 11/10-i(sqrt(39))/(10)

Find the zeros of 5x^2-11x+8=0

The zeros of 5x^2-11x+8=0 are x= 11/10+i(sqrt(39))/(10),x= 11/10-i(sqrt(39))/(10)

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5 x^{2} - 11 x + 8 = 0

x = \frac{11}{10} + i \frac{39}{10}, x = \frac{11}{10} - i \frac{39}{10}

Solution steps

5 x^{2} - 11 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 8}{2 \cdot 5}

Simplify

(- 11)^{2} - 4 \cdot 5 \cdot 8 : 39 i

x_{1, 2} = \frac{- ( - 11 ) \pm 39 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 39 i}{2 \cdot 5}, x_{2} = \frac{- ( - 11 ) - 39 i}{2 \cdot 5}

x = \frac{- ( - 11 ) + 39 i}{2 \cdot 5} : \frac{11}{10} + i \frac{39}{10}

x = \frac{- ( - 11 ) - 39 i}{2 \cdot 5} : \frac{11}{10} - i \frac{39}{10}

The solutions to the quadratic equation are:

x = \frac{11}{10} + i \frac{39}{10}, x = \frac{11}{10} - i \frac{39}{10}

t^{2} + 24 t + 144

r - 2 > 11

7 x + 10 = - 4 x - 23

b^{2} - 6 b - 16 = 0

What are the solutions to the equation 5x^2-11x+8=0 ?
The solutions to the equation 5x^2-11x+8=0 are x= 11/10+i(sqrt(39))/(10),x= 11/10-i(sqrt(39))/(10)
Find the zeros of 5x^2-11x+8=0
The zeros of 5x^2-11x+8=0 are x= 11/10+i(sqrt(39))/(10),x= 11/10-i(sqrt(39))/(10)