What are the solutions to the equation (x)=4x^2-9x+16 ?

The solutions to the equation (x)=4x^2-9x+16 are x= 5/4+i(sqrt(39))/4 ,x= 5/4-i(sqrt(39))/4

Find the zeros of (x)=4x^2-9x+16

The zeros of (x)=4x^2-9x+16 are x= 5/4+i(sqrt(39))/4 ,x= 5/4-i(sqrt(39))/4

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(x) = 4 x^{2} - 9 x + 16

x = \frac{5}{4} + i \frac{39}{4}, x = \frac{5}{4} - i \frac{39}{4}

Solution steps

(x) = 4 x^{2} - 9 x + 16

Refine

x = 4 x^{2} - 9 x + 16

Switch sides

4 x^{2} - 9 x + 16 = x

Move

x

to the left side

4 x^{2} - 10 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

Simplify

(- 10)^{2} - 4 \cdot 4 \cdot 16 : 239 i

x_{1, 2} = \frac{- ( - 10 ) \pm 2 39 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 2 39 i}{2 \cdot 4}, x_{2} = \frac{- ( - 10 ) - 2 39 i}{2 \cdot 4}

x = \frac{- ( - 10 ) + 2 39 i}{2 \cdot 4} : \frac{5}{4} + i \frac{39}{4}

x = \frac{- ( - 10 ) - 2 39 i}{2 \cdot 4} : \frac{5}{4} - i \frac{39}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{39}{4}, x = \frac{5}{4} - i \frac{39}{4}

x^{2} - 3 x + 2 = 0, (1, - 1, 2, - 2)

2 m^{2} + 8 m + 8 = 0

- 5 t^{2} + 9 t + 3 = 0

2 x^{2} + 20 x - 8 = 0

321.0069 x^{2} - 5130.1778 x + 14682.6031 = 0

What are the solutions to the equation (x)=4x^2-9x+16 ?
The solutions to the equation (x)=4x^2-9x+16 are x= 5/4+i(sqrt(39))/4 ,x= 5/4-i(sqrt(39))/4
Find the zeros of (x)=4x^2-9x+16
The zeros of (x)=4x^2-9x+16 are x= 5/4+i(sqrt(39))/4 ,x= 5/4-i(sqrt(39))/4