What are the solutions to the equation 4q^2-2q=-65 ?

The solutions to the equation 4q^2-2q=-65 are q= 1/4+i(sqrt(259))/4 ,q= 1/4-i(sqrt(259))/4

Find the zeros of 4q^2-2q=-65

The zeros of 4q^2-2q=-65 are q= 1/4+i(sqrt(259))/4 ,q= 1/4-i(sqrt(259))/4

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4 q^{2} - 2 q = - 65

q = \frac{1}{4} + i \frac{259}{4}, q = \frac{1}{4} - i \frac{259}{4}

Solution steps

4 q^{2} - 2 q = - 65

Move

65

to the left side

4 q^{2} - 2 q + 65 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 4 \cdot 65}{2 \cdot 4}

Simplify

(- 2)^{2} - 4 \cdot 4 \cdot 65 : 2259 i

q_{1, 2} = \frac{- ( - 2 ) \pm 2 259 i}{2 \cdot 4}

Separate the solutions

q_{1} = \frac{- ( - 2 ) + 2 259 i}{2 \cdot 4}, q_{2} = \frac{- ( - 2 ) - 2 259 i}{2 \cdot 4}

q = \frac{- ( - 2 ) + 2 259 i}{2 \cdot 4} : \frac{1}{4} + i \frac{259}{4}

q = \frac{- ( - 2 ) - 2 259 i}{2 \cdot 4} : \frac{1}{4} - i \frac{259}{4}

The solutions to the quadratic equation are:

q = \frac{1}{4} + i \frac{259}{4}, q = \frac{1}{4} - i \frac{259}{4}

(x - 15)^{2} = 225

4 (x - 6)^{2} - 27 = - 23

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

4 r^{2} + 1 = 325

x, x^{2} + x y + y^{3} = 0

What are the solutions to the equation 4q^2-2q=-65 ?
The solutions to the equation 4q^2-2q=-65 are q= 1/4+i(sqrt(259))/4 ,q= 1/4-i(sqrt(259))/4
Find the zeros of 4q^2-2q=-65
The zeros of 4q^2-2q=-65 are q= 1/4+i(sqrt(259))/4 ,q= 1/4-i(sqrt(259))/4