What are the solutions to the equation 2x^2-15x+32=0 ?

The solutions to the equation 2x^2-15x+32=0 are x= 15/4+i(sqrt(31))/4 ,x= 15/4-i(sqrt(31))/4

Find the zeros of 2x^2-15x+32=0

The zeros of 2x^2-15x+32=0 are x= 15/4+i(sqrt(31))/4 ,x= 15/4-i(sqrt(31))/4

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2 x^{2} - 15 x + 32 = 0

x = \frac{15}{4} + i \frac{31}{4}, x = \frac{15}{4} - i \frac{31}{4}

Solution steps

2 x^{2} - 15 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

Simplify

(- 15)^{2} - 4 \cdot 2 \cdot 32 : 31 i

x_{1, 2} = \frac{- ( - 15 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 31 i}{2 \cdot 2}, x_{2} = \frac{- ( - 15 ) - 31 i}{2 \cdot 2}

x = \frac{- ( - 15 ) + 31 i}{2 \cdot 2} : \frac{15}{4} + i \frac{31}{4}

x = \frac{- ( - 15 ) - 31 i}{2 \cdot 2} : \frac{15}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{15}{4} + i \frac{31}{4}, x = \frac{15}{4} - i \frac{31}{4}

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What are the solutions to the equation 2x^2-15x+32=0 ?
The solutions to the equation 2x^2-15x+32=0 are x= 15/4+i(sqrt(31))/4 ,x= 15/4-i(sqrt(31))/4
Find the zeros of 2x^2-15x+32=0
The zeros of 2x^2-15x+32=0 are x= 15/4+i(sqrt(31))/4 ,x= 15/4-i(sqrt(31))/4