What are the solutions to the equation 2x^2-5x+32=7 ?

The solutions to the equation 2x^2-5x+32=7 are x= 5/4+i(5sqrt(7))/4 ,x= 5/4-i(5sqrt(7))/4

Find the zeros of 2x^2-5x+32=7

The zeros of 2x^2-5x+32=7 are x= 5/4+i(5sqrt(7))/4 ,x= 5/4-i(5sqrt(7))/4

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2 x^{2} - 5 x + 32 = 7

x = \frac{5}{4} + i \frac{5 7}{4}, x = \frac{5}{4} - i \frac{5 7}{4}

Solution steps

2 x^{2} - 5 x + 32 = 7

Move

7

to the left side

2 x^{2} - 5 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 25}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 25 : 57 i

x_{1, 2} = \frac{- ( - 5 ) \pm 5 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 5 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 5 7 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 5 7 i}{2 \cdot 2} : \frac{5}{4} + i \frac{5 7}{4}

x = \frac{- ( - 5 ) - 5 7 i}{2 \cdot 2} : \frac{5}{4} - i \frac{5 7}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{5 7}{4}, x = \frac{5}{4} - i \frac{5 7}{4}

3 x^{2} + 24 x + 57 = 9

2 x^{2} - 0.45 x - 3 = 0

15 c \cdot (a + b) + 8 \cdot (b + a)

- 5 x^{3} + 6 x^{2} + 15 x - 18

x^{4} - x^{2} - 2 x - 1

What are the solutions to the equation 2x^2-5x+32=7 ?
The solutions to the equation 2x^2-5x+32=7 are x= 5/4+i(5sqrt(7))/4 ,x= 5/4-i(5sqrt(7))/4
Find the zeros of 2x^2-5x+32=7
The zeros of 2x^2-5x+32=7 are x= 5/4+i(5sqrt(7))/4 ,x= 5/4-i(5sqrt(7))/4