What are the solutions to the equation 4z(2z+1)=-6z-3 ?

The solutions to the equation 4z(2z+1)=-6z-3 are z=-1/2 ,z=-3/4

Find the zeros of 4z(2z+1)=-6z-3

The zeros of 4z(2z+1)=-6z-3 are z=-1/2 ,z=-3/4

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4 z (2 z + 1) = - 6 z - 3

z = - \frac{1}{2}, z = - \frac{3}{4}

Decimal Notation

z = - 0.5, z = - 0.75

Solution steps

4 z (2 z + 1) = - 6 z - 3

Expand

4 z (2 z + 1) : 8 z^{2} + 4 z

8 z^{2} + 4 z = - 6 z - 3

Move

3

to the left side

8 z^{2} + 4 z + 3 = - 6 z

Move

6 z

to the left side

8 z^{2} + 10 z + 3 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

1 0^{2} - 4 \cdot 8 \cdot 3 = 2

z_{1, 2} = \frac{- 10 \pm 2}{2 \cdot 8}

Separate the solutions

z_{1} = \frac{- 10 + 2}{2 \cdot 8}, z_{2} = \frac{- 10 - 2}{2 \cdot 8}

z = \frac{- 10 + 2}{2 \cdot 8} : - \frac{1}{2}

z = \frac{- 10 - 2}{2 \cdot 8} : - \frac{3}{4}

The solutions to the quadratic equation are:

z = - \frac{1}{2}, z = - \frac{3}{4}

x (6 x - 1) - 3 (6 x - 1)

(\frac{1}{81})^{\frac{1}{2}} \cdot (\frac{1}{81})^{- \frac{3}{4}}

(\frac{( 2 ^{2} 4 ^{2} 2 )}{4 \cdot 2 ^{3}})^{2}

\frac{10 q ^{5} w ^{7}}{2 w ^{3}} \cdot \frac{4 ( q ^{6} ) ^{2}}{w ^{- 5}}

(x^{2} - 4) = 0

What are the solutions to the equation 4z(2z+1)=-6z-3 ?
The solutions to the equation 4z(2z+1)=-6z-3 are z=-1/2 ,z=-3/4
Find the zeros of 4z(2z+1)=-6z-3
The zeros of 4z(2z+1)=-6z-3 are z=-1/2 ,z=-3/4