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

The solutions to the equation 2x^2-5x=-7 are x= 5/4+i(sqrt(31))/4 ,x= 5/4-i(sqrt(31))/4

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

The zeros of 2x^2-5x=-7 are x= 5/4+i(sqrt(31))/4 ,x= 5/4-i(sqrt(31))/4

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

x = \frac{5}{4} + i \frac{31}{4}, x = \frac{5}{4} - i \frac{31}{4}

Solution steps

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

Move

7

to the left side

2 x^{2} - 5 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 7 : 31 i

x_{1, 2} = \frac{- ( - 5 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 31 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 31 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 31 i}{2 \cdot 2} : \frac{5}{4} + i \frac{31}{4}

x = \frac{- ( - 5 ) - 31 i}{2 \cdot 2} : \frac{5}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{31}{4}, x = \frac{5}{4} - i \frac{31}{4}

- \frac{4}{2} - \frac{8}{3}

27 - 5 x^{2} = - 53

9 q + 2 > - 3 q + 1

x (x + 1) + 2 (x + 1) + x (x - 3) + 2 (x - 3)

\frac{15 x + 12}{3} = 5 x + 4

What are the solutions to the equation 2x^2-5x=-7 ?
The solutions to the equation 2x^2-5x=-7 are x= 5/4+i(sqrt(31))/4 ,x= 5/4-i(sqrt(31))/4
Find the zeros of 2x^2-5x=-7
The zeros of 2x^2-5x=-7 are x= 5/4+i(sqrt(31))/4 ,x= 5/4-i(sqrt(31))/4