What are the solutions to the equation (2x-3)(2x-1)=-4 ?

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

Find the zeros of (2x-3)(2x-1)=-4

The zeros of (2x-3)(2x-1)=-4 are x=1+i(sqrt(3))/2 ,x=1-i(sqrt(3))/2

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(2 x - 3) (2 x - 1) = - 4

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

Solution steps

(2 x - 3) (2 x - 1) = - 4

Expand

(2 x - 3) (2 x - 1) : 4 x^{2} - 8 x + 3

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

Move

4

to the left side

4 x^{2} - 8 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 \cdot 7}{2 \cdot 4}

Simplify

(- 8)^{2} - 4 \cdot 4 \cdot 7 : 43 i

x_{1, 2} = \frac{- ( - 8 ) \pm 4 3 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 3 i}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 4 3 i}{2 \cdot 4}

x = \frac{- ( - 8 ) + 4 3 i}{2 \cdot 4} : 1 + i \frac{3}{2}

x = \frac{- ( - 8 ) - 4 3 i}{2 \cdot 4} : 1 - i \frac{3}{2}

The solutions to the quadratic equation are:

x = 1 + i \frac{3}{2}, x = 1 - i \frac{3}{2}

5 x^{2} = 35

- 9 - 9

- \frac{3}{5 ^{2}}

\frac{20 - 17 i}{2 - 7 i}

2 x^{2} + 16 x + 30

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