What are the solutions to the equation 4j^2=-32+66j ?

The solutions to the equation 4j^2=-32+66j are j=16,j= 1/2

Find the zeros of 4j^2=-32+66j

The zeros of 4j^2=-32+66j are j=16,j= 1/2

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4 j^{2} = - 32 + 66 j

j = 16, j = \frac{1}{2}

Decimal Notation

j = 16, j = 0.5

Solution steps

4 j^{2} = - 32 + 66 j

Move

66 j

to the left side

4 j^{2} - 66 j = - 32

Move

32

to the left side

4 j^{2} - 66 j + 32 = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- ( - 66 ) \pm ( - 66 ) ^{2} - 4 \cdot 4 \cdot 32}{2 \cdot 4}

(- 66)^{2} - 4 \cdot 4 \cdot 32 = 62

j_{1, 2} = \frac{- ( - 66 ) \pm 62}{2 \cdot 4}

Separate the solutions

j_{1} = \frac{- ( - 66 ) + 62}{2 \cdot 4}, j_{2} = \frac{- ( - 66 ) - 62}{2 \cdot 4}

j = \frac{- ( - 66 ) + 62}{2 \cdot 4} : 16

j = \frac{- ( - 66 ) - 62}{2 \cdot 4} : \frac{1}{2}

The solutions to the quadratic equation are:

j = 16, j = \frac{1}{2}

∣ 2 x + 1 ∣ \geq 2 + x

5 y + (- 5) = 10

y^{2} - 10 y + 26 = 0

x (x - 1) - 5 (x - 2) = 2

6 x - x^{2} = 8

What are the solutions to the equation 4j^2=-32+66j ?
The solutions to the equation 4j^2=-32+66j are j=16,j= 1/2
Find the zeros of 4j^2=-32+66j
The zeros of 4j^2=-32+66j are j=16,j= 1/2