What are the solutions to the equation 4x^2=32x-67 ?

The solutions to the equation 4x^2=32x-67 are x=4+i(sqrt(3))/2 ,x=4-i(sqrt(3))/2

Find the zeros of 4x^2=32x-67

The zeros of 4x^2=32x-67 are x=4+i(sqrt(3))/2 ,x=4-i(sqrt(3))/2

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4 x^{2} = 32 x - 67

x = 4 + i \frac{3}{2}, x = 4 - i \frac{3}{2}

Solution steps

4 x^{2} = 32 x - 67

Move

67

to the left side

4 x^{2} + 67 = 32 x

Move

32 x

to the left side

4 x^{2} + 67 - 32 x = 0

Write in the standard form

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

4 x^{2} - 32 x + 67 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 4 \cdot 67}{2 \cdot 4}

Simplify

(- 32)^{2} - 4 \cdot 4 \cdot 67 : 43 i

x_{1, 2} = \frac{- ( - 32 ) \pm 4 3 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 4 3 i}{2 \cdot 4}, x_{2} = \frac{- ( - 32 ) - 4 3 i}{2 \cdot 4}

x = \frac{- ( - 32 ) + 4 3 i}{2 \cdot 4} : 4 + i \frac{3}{2}

x = \frac{- ( - 32 ) - 4 3 i}{2 \cdot 4} : 4 - i \frac{3}{2}

The solutions to the quadratic equation are:

x = 4 + i \frac{3}{2}, x = 4 - i \frac{3}{2}

(n + 4)^{2} = 2 n^{2} + 14 n + 24

2 x^{2} - 3 x - 104 = 0

- x (3 - x) + 2 = 0

w^{2} - \frac{1}{15} w - \frac{2}{15} = 0

3 x^{2} + 6 x - 1

What are the solutions to the equation 4x^2=32x-67 ?
The solutions to the equation 4x^2=32x-67 are x=4+i(sqrt(3))/2 ,x=4-i(sqrt(3))/2
Find the zeros of 4x^2=32x-67
The zeros of 4x^2=32x-67 are x=4+i(sqrt(3))/2 ,x=4-i(sqrt(3))/2