What are the solutions to the equation 0=2(2x^2-3x+2) ?

The solutions to the equation 0=2(2x^2-3x+2) are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4

Find the zeros of 0=2(2x^2-3x+2)

The zeros of 0=2(2x^2-3x+2) are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4

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

x = \frac{3}{4} + i \frac{7}{4}, x = \frac{3}{4} - i \frac{7}{4}

Solution steps

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

Expand

2 (2 x^{2} - 3 x + 2) : 4 x^{2} - 6 x + 4

0 = 4 x^{2} - 6 x + 4

Switch sides

4 x^{2} - 6 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

Simplify

(- 6)^{2} - 4 \cdot 4 \cdot 4 : 27 i

x_{1, 2} = \frac{- ( - 6 ) \pm 2 7 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 7 i}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 2 7 i}{2 \cdot 4}

x = \frac{- ( - 6 ) + 2 7 i}{2 \cdot 4} : \frac{3}{4} + i \frac{7}{4}

x = \frac{- ( - 6 ) - 2 7 i}{2 \cdot 4} : \frac{3}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{7}{4}, x = \frac{3}{4} - i \frac{7}{4}

10 = 2 + 23 t - 16 t^{2}

x^{2} - x - 10 = x + 5

x^{2} - 6 x - 6 = 0

3 x - 5 = x^{2} - 2 x + 1

4 w^{2} = - w + 3

What are the solutions to the equation 0=2(2x^2-3x+2) ?
The solutions to the equation 0=2(2x^2-3x+2) are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4
Find the zeros of 0=2(2x^2-3x+2)
The zeros of 0=2(2x^2-3x+2) are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4