What are the solutions to the equation 2x=4x^2-5x+50 ?

The solutions to the equation 2x=4x^2-5x+50 are x= 7/8+i(sqrt(751))/8 ,x= 7/8-i(sqrt(751))/8

Find the zeros of 2x=4x^2-5x+50

The zeros of 2x=4x^2-5x+50 are x= 7/8+i(sqrt(751))/8 ,x= 7/8-i(sqrt(751))/8

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2 x = 4 x^{2} - 5 x + 50

x = \frac{7}{8} + i \frac{751}{8}, x = \frac{7}{8} - i \frac{751}{8}

Solution steps

2 x = 4 x^{2} - 5 x + 50

Switch sides

4 x^{2} - 5 x + 50 = 2 x

Move

2 x

to the left side

4 x^{2} - 7 x + 50 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 50}{2 \cdot 4}

Simplify

(- 7)^{2} - 4 \cdot 4 \cdot 50 : 751 i

x_{1, 2} = \frac{- ( - 7 ) \pm 751 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 751 i}{2 \cdot 4}, x_{2} = \frac{- ( - 7 ) - 751 i}{2 \cdot 4}

x = \frac{- ( - 7 ) + 751 i}{2 \cdot 4} : \frac{7}{8} + i \frac{751}{8}

x = \frac{- ( - 7 ) - 751 i}{2 \cdot 4} : \frac{7}{8} - i \frac{751}{8}

The solutions to the quadratic equation are:

x = \frac{7}{8} + i \frac{751}{8}, x = \frac{7}{8} - i \frac{751}{8}

v^{2} + 10 v + 16 = 0

4 x^{2} - 5 x - 21 = 0

n^{2} - 9 = 0

4 x^{2} + x = - 3

(x - 3)^{2} + 7 = 23

What are the solutions to the equation 2x=4x^2-5x+50 ?
The solutions to the equation 2x=4x^2-5x+50 are x= 7/8+i(sqrt(751))/8 ,x= 7/8-i(sqrt(751))/8
Find the zeros of 2x=4x^2-5x+50
The zeros of 2x=4x^2-5x+50 are x= 7/8+i(sqrt(751))/8 ,x= 7/8-i(sqrt(751))/8