What are the solutions to the equation 2j^2-3j=-1 ?

The solutions to the equation 2j^2-3j=-1 are j=1,j= 1/2

Find the zeros of 2j^2-3j=-1

The zeros of 2j^2-3j=-1 are j=1,j= 1/2

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2 j^{2} - 3 j = - 1

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

Decimal Notation

j = 1, j = 0.5

Solution steps

2 j^{2} - 3 j = - 1

Move

1

to the left side

2 j^{2} - 3 j + 1 = 0

Solve with the quadratic formula

j_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

j_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 2}

Separate the solutions

j_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 2}, j_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 2}

j = \frac{- ( - 3 ) + 1}{2 \cdot 2} : 1

j = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

The solutions to the quadratic equation are:

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

14 y^{2} - 112 y = 0

- 3 x^{2} + 5 x - 6 = 0

10 x^{2} + 2 x = 9

x, x^{2} - x = y^{2} - y

x^{2} + 72 = 12 x

What are the solutions to the equation 2j^2-3j=-1 ?
The solutions to the equation 2j^2-3j=-1 are j=1,j= 1/2
Find the zeros of 2j^2-3j=-1
The zeros of 2j^2-3j=-1 are j=1,j= 1/2