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

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

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

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

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2 x^{2} + 7 = 5 x

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

Solution steps

2 x^{2} + 7 = 5 x

Move

5 x

to the left side

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

Write in the standard form

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

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}

2 x^{2} + 7 = 8 x

3 x^{2} + 12 x - 15 x = 0

- x^{2} + 11 x - 10 = 0

(w + 7) (w - 2) = 0

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

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