What are the solutions to the equation 2x^2+11=9x ?

The solutions to the equation 2x^2+11=9x are x= 9/4+i(sqrt(7))/4 ,x= 9/4-i(sqrt(7))/4

Find the zeros of 2x^2+11=9x

The zeros of 2x^2+11=9x are x= 9/4+i(sqrt(7))/4 ,x= 9/4-i(sqrt(7))/4

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2 x^{2} + 11 = 9 x

x = \frac{9}{4} + i \frac{7}{4}, x = \frac{9}{4} - i \frac{7}{4}

Solution steps

2 x^{2} + 11 = 9 x

Move

9 x

to the left side

2 x^{2} + 11 - 9 x = 0

Write in the standard form

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

2 x^{2} - 9 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 11}{2 \cdot 2}

Simplify

(- 9)^{2} - 4 \cdot 2 \cdot 11 : 7 i

x_{1, 2} = \frac{- ( - 9 ) \pm 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 7 i}{2 \cdot 2}

x = \frac{- ( - 9 ) + 7 i}{2 \cdot 2} : \frac{9}{4} + i \frac{7}{4}

x = \frac{- ( - 9 ) - 7 i}{2 \cdot 2} : \frac{9}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{9}{4} + i \frac{7}{4}, x = \frac{9}{4} - i \frac{7}{4}

x, x^{2} - 4 x + 4 = 0

x^{2} - 6 x - 7

x^{2} - 32 x + 4 = 0

5 = 4 n^{2} + n + 5

2 x^{2} + x + 10 = 5 - 4 x - x^{2}

What are the solutions to the equation 2x^2+11=9x ?
The solutions to the equation 2x^2+11=9x are x= 9/4+i(sqrt(7))/4 ,x= 9/4-i(sqrt(7))/4
Find the zeros of 2x^2+11=9x
The zeros of 2x^2+11=9x are x= 9/4+i(sqrt(7))/4 ,x= 9/4-i(sqrt(7))/4