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

The solutions to the equation 2x^2-5x=-5 are x= 5/4+i(sqrt(15))/4 ,x= 5/4-i(sqrt(15))/4

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

The zeros of 2x^2-5x=-5 are x= 5/4+i(sqrt(15))/4 ,x= 5/4-i(sqrt(15))/4

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

x = \frac{5}{4} + i \frac{15}{4}, x = \frac{5}{4} - i \frac{15}{4}

Solution steps

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

Move

5

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 5 : 15 i

x_{1, 2} = \frac{- ( - 5 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 15 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 15 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 15 i}{2 \cdot 2} : \frac{5}{4} + i \frac{15}{4}

x = \frac{- ( - 5 ) - 15 i}{2 \cdot 2} : \frac{5}{4} - i \frac{15}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{15}{4}, x = \frac{5}{4} - i \frac{15}{4}

x^{2} - 5 x + 9 = 3

x^{2} + 4 x - 3

x^{2} - 6 = - 5 x

- 4.9 t^{2} + 25 t = 0

- 4 x^{2} - 72 x - 68 = 0

What are the solutions to the equation 2x^2-5x=-5 ?
The solutions to the equation 2x^2-5x=-5 are x= 5/4+i(sqrt(15))/4 ,x= 5/4-i(sqrt(15))/4
Find the zeros of 2x^2-5x=-5
The zeros of 2x^2-5x=-5 are x= 5/4+i(sqrt(15))/4 ,x= 5/4-i(sqrt(15))/4