What are the solutions to the equation 13-6x=(2x-5)^2+3 ?

The solutions to the equation 13-6x=(2x-5)^2+3 are x= 7/4+i(sqrt(11))/4 ,x= 7/4-i(sqrt(11))/4

Find the zeros of 13-6x=(2x-5)^2+3

The zeros of 13-6x=(2x-5)^2+3 are x= 7/4+i(sqrt(11))/4 ,x= 7/4-i(sqrt(11))/4

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13 - 6 x = (2 x - 5)^{2} + 3

x = \frac{7}{4} + i \frac{11}{4}, x = \frac{7}{4} - i \frac{11}{4}

Solution steps

13 - 6 x = (2 x - 5)^{2} + 3

Expand

(2 x - 5)^{2} + 3 : 4 x^{2} - 20 x + 28

13 - 6 x = 4 x^{2} - 20 x + 28

Switch sides

4 x^{2} - 20 x + 28 = 13 - 6 x

Move

6 x

to the left side

4 x^{2} - 14 x + 28 = 13

Move

13

to the left side

4 x^{2} - 14 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 4 \cdot 15}{2 \cdot 4}

Simplify

(- 14)^{2} - 4 \cdot 4 \cdot 15 : 211 i

x_{1, 2} = \frac{- ( - 14 ) \pm 2 11 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 2 11 i}{2 \cdot 4}, x_{2} = \frac{- ( - 14 ) - 2 11 i}{2 \cdot 4}

x = \frac{- ( - 14 ) + 2 11 i}{2 \cdot 4} : \frac{7}{4} + i \frac{11}{4}

x = \frac{- ( - 14 ) - 2 11 i}{2 \cdot 4} : \frac{7}{4} - i \frac{11}{4}

The solutions to the quadratic equation are:

x = \frac{7}{4} + i \frac{11}{4}, x = \frac{7}{4} - i \frac{11}{4}

x, - \frac{1}{y} = x - 4 x^{2}

0 = - 4.9 t^{2} + 4.25 t + 100

m^{2} + 17 m + 72 = 0

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

9 m^{2} - 1 = 0

What are the solutions to the equation 13-6x=(2x-5)^2+3 ?
The solutions to the equation 13-6x=(2x-5)^2+3 are x= 7/4+i(sqrt(11))/4 ,x= 7/4-i(sqrt(11))/4
Find the zeros of 13-6x=(2x-5)^2+3
The zeros of 13-6x=(2x-5)^2+3 are x= 7/4+i(sqrt(11))/4 ,x= 7/4-i(sqrt(11))/4