What are the solutions to the equation 2x^2+5=3x ?

The solutions to the equation 2x^2+5=3x are x= 3/4+i(sqrt(31))/4 ,x= 3/4-i(sqrt(31))/4

Find the zeros of 2x^2+5=3x

The zeros of 2x^2+5=3x are x= 3/4+i(sqrt(31))/4 ,x= 3/4-i(sqrt(31))/4

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2 x^{2} + 5 = 3 x

x = \frac{3}{4} + i \frac{31}{4}, x = \frac{3}{4} - i \frac{31}{4}

Solution steps

2 x^{2} + 5 = 3 x

Move

3 x

to the left side

2 x^{2} + 5 - 3 x = 0

Write in the standard form

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

2 x^{2} - 3 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 5 : 31 i

x_{1, 2} = \frac{- ( - 3 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 31 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 31 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 31 i}{2 \cdot 2} : \frac{3}{4} + i \frac{31}{4}

x = \frac{- ( - 3 ) - 31 i}{2 \cdot 2} : \frac{3}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{31}{4}, x = \frac{3}{4} - i \frac{31}{4}

2 x^{2} + 5 = 77

a, b^{2} = c^{2} - a^{2}

x^{2} - 5 x = 2 (4 x - 11)

x^{2} - 24 = 5 x

4 x^{2} - 12 x + 9

What are the solutions to the equation 2x^2+5=3x ?
The solutions to the equation 2x^2+5=3x are x= 3/4+i(sqrt(31))/4 ,x= 3/4-i(sqrt(31))/4
Find the zeros of 2x^2+5=3x
The zeros of 2x^2+5=3x are x= 3/4+i(sqrt(31))/4 ,x= 3/4-i(sqrt(31))/4