What are the solutions to the equation z(2z-3)=+4 ?

The solutions to the equation z(2z-3)=+4 are z=(3+sqrt(41))/4 ,z=(3-sqrt(41))/4

Find the zeros of z(2z-3)=+4

The zeros of z(2z-3)=+4 are z=(3+sqrt(41))/4 ,z=(3-sqrt(41))/4

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z (2 z - 3) = + 4

z = \frac{3 + 41}{4}, z = \frac{3 - 41}{4}

Decimal Notation

z = 2.35078 \dots, z = - 0.85078 \dots

Solution steps

z (2 z - 3) = 4

Expand

z (2 z - 3) : 2 z^{2} - 3 z

2 z^{2} - 3 z = 4

Move

4

to the left side

2 z^{2} - 3 z - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 2 (- 4) = 41

z_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 \cdot 2}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 41}{2 \cdot 2}, z_{2} = \frac{- ( - 3 ) - 41}{2 \cdot 2}

z = \frac{- ( - 3 ) + 41}{2 \cdot 2} : \frac{3 + 41}{4}

z = \frac{- ( - 3 ) - 41}{2 \cdot 2} : \frac{3 - 41}{4}

The solutions to the quadratic equation are:

z = \frac{3 + 41}{4}, z = \frac{3 - 41}{4}

\frac{( 9 ^{- 3} )}{3 ^{- 8}}

- 10 < - 5 + 9 m \leq - 8

- \frac{3}{4} x \geq - \frac{3}{8} + \frac{2}{3} x

- 3 m - 6 = 18

2 x + 5 < - 13

What are the solutions to the equation z(2z-3)=+4 ?
The solutions to the equation z(2z-3)=+4 are z=(3+sqrt(41))/4 ,z=(3-sqrt(41))/4
Find the zeros of z(2z-3)=+4
The zeros of z(2z-3)=+4 are z=(3+sqrt(41))/4 ,z=(3-sqrt(41))/4