What are the solutions to the equation 2x^2-3x=-2 ?

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

Find the zeros of 2x^2-3x=-2

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

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

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

Solution steps

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

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 2 : 7 i

x_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 2}

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

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

115 + 12.25 t - 4.9 t^{2}

(m - 3) x^{2} + (m - 7) x + 2 = 0

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

5 x^{2} + 7 = 27

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

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