What are the solutions to the equation 2x^2+6=5x ?

The solutions to the equation 2x^2+6=5x are x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4

Find the zeros of 2x^2+6=5x

The zeros of 2x^2+6=5x are x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4

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2 x^{2} + 6 = 5 x

x = \frac{5}{4} + i \frac{23}{4}, x = \frac{5}{4} - i \frac{23}{4}

Solution steps

2 x^{2} + 6 = 5 x

Move

5 x

to the left side

2 x^{2} + 6 - 5 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 6 : 23 i

x_{1, 2} = \frac{- ( - 5 ) \pm 23 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 23 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 23 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 23 i}{2 \cdot 2} : \frac{5}{4} + i \frac{23}{4}

x = \frac{- ( - 5 ) - 23 i}{2 \cdot 2} : \frac{5}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{23}{4}, x = \frac{5}{4} - i \frac{23}{4}

- 31 - s > 93

26 x + 21 = 23 x - 9

8 - \frac{4 + 3 m}{5} = \frac{3 m}{10}

\frac{3 + 90}{9}

What are the solutions to the equation 2x^2+6=5x ?
The solutions to the equation 2x^2+6=5x are x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4
Find the zeros of 2x^2+6=5x
The zeros of 2x^2+6=5x are x= 5/4+i(sqrt(23))/4 ,x= 5/4-i(sqrt(23))/4