What are the solutions to the equation 4(-3x+2x^2)+9=0 ?

The solutions to the equation 4(-3x+2x^2)+9=0 are x= 3/4+i 3/4 ,x= 3/4-i 3/4

Find the zeros of 4(-3x+2x^2)+9=0

The zeros of 4(-3x+2x^2)+9=0 are x= 3/4+i 3/4 ,x= 3/4-i 3/4

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4 (- 3 x + 2 x^{2}) + 9 = 0

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

Solution steps

4 (- 3 x + 2 x^{2}) + 9 = 0

Expand

4 (- 3 x + 2 x^{2}) + 9 : - 12 x + 8 x^{2} + 9

- 12 x + 8 x^{2} + 9 = 0

Write in the standard form

a x^{2} + b x + c = 0

8 x^{2} - 12 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

Simplify

(- 12)^{2} - 4 \cdot 8 \cdot 9 : 12 i

x_{1, 2} = \frac{- ( - 12 ) \pm 12 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 12 i}{2 \cdot 8}, x_{2} = \frac{- ( - 12 ) - 12 i}{2 \cdot 8}

x = \frac{- ( - 12 ) + 12 i}{2 \cdot 8} : \frac{3}{4} + i \frac{3}{4}

x = \frac{- ( - 12 ) - 12 i}{2 \cdot 8} : \frac{3}{4} - i \frac{3}{4}

The solutions to the quadratic equation are:

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

8 x^{2} = 6 + 22 x

x^{2} + 5 x - 176 = 0

s^{2} + s - 12 = 0

10 x^{2} + 20 x + 6260 = 0

11 r^{2} - 5 r - 12 = 7

What are the solutions to the equation 4(-3x+2x^2)+9=0 ?
The solutions to the equation 4(-3x+2x^2)+9=0 are x= 3/4+i 3/4 ,x= 3/4-i 3/4
Find the zeros of 4(-3x+2x^2)+9=0
The zeros of 4(-3x+2x^2)+9=0 are x= 3/4+i 3/4 ,x= 3/4-i 3/4