What are the solutions to the equation 4x^2-3x+9=2x+1 ?

The solutions to the equation 4x^2-3x+9=2x+1 are x= 5/8+i(sqrt(103))/8 ,x= 5/8-i(sqrt(103))/8

Find the zeros of 4x^2-3x+9=2x+1

The zeros of 4x^2-3x+9=2x+1 are x= 5/8+i(sqrt(103))/8 ,x= 5/8-i(sqrt(103))/8

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4 x^{2} - 3 x + 9 = 2 x + 1

x = \frac{5}{8} + i \frac{103}{8}, x = \frac{5}{8} - i \frac{103}{8}

Solution steps

4 x^{2} - 3 x + 9 = 2 x + 1

Move

1

to the left side

4 x^{2} - 3 x + 8 = 2 x

Move

2 x

to the left side

4 x^{2} - 5 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 8}{2 \cdot 4}

Simplify

(- 5)^{2} - 4 \cdot 4 \cdot 8 : 103 i

x_{1, 2} = \frac{- ( - 5 ) \pm 103 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 103 i}{2 \cdot 4}, x_{2} = \frac{- ( - 5 ) - 103 i}{2 \cdot 4}

x = \frac{- ( - 5 ) + 103 i}{2 \cdot 4} : \frac{5}{8} + i \frac{103}{8}

x = \frac{- ( - 5 ) - 103 i}{2 \cdot 4} : \frac{5}{8} - i \frac{103}{8}

The solutions to the quadratic equation are:

x = \frac{5}{8} + i \frac{103}{8}, x = \frac{5}{8} - i \frac{103}{8}

15 \cdot 8

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

7 x^{2} + 22 x - 5 = 0

5 \leq \frac{j}{3} + 6

\frac{170 - 125}{125}

What are the solutions to the equation 4x^2-3x+9=2x+1 ?
The solutions to the equation 4x^2-3x+9=2x+1 are x= 5/8+i(sqrt(103))/8 ,x= 5/8-i(sqrt(103))/8
Find the zeros of 4x^2-3x+9=2x+1
The zeros of 4x^2-3x+9=2x+1 are x= 5/8+i(sqrt(103))/8 ,x= 5/8-i(sqrt(103))/8