What are the solutions to the equation 18z^2-31z+12=-2z^2 ?

The solutions to the equation 18z^2-31z+12=-2z^2 are z= 4/5 ,z= 3/4

Find the zeros of 18z^2-31z+12=-2z^2

The zeros of 18z^2-31z+12=-2z^2 are z= 4/5 ,z= 3/4

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18 z^{2} - 31 z + 12 = - 2 z^{2}

z = \frac{4}{5}, z = \frac{3}{4}

Decimal Notation

z = 0.8, z = 0.75

Solution steps

18 z^{2} - 31 z + 12 = - 2 z^{2}

Move

2 z^{2}

to the left side

20 z^{2} - 31 z + 12 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 31 ) \pm ( - 31 ) ^{2} - 4 \cdot 20 \cdot 12}{2 \cdot 20}

(- 31)^{2} - 4 \cdot 20 \cdot 12 = 1

z_{1, 2} = \frac{- ( - 31 ) \pm 1}{2 \cdot 20}

Separate the solutions

z_{1} = \frac{- ( - 31 ) + 1}{2 \cdot 20}, z_{2} = \frac{- ( - 31 ) - 1}{2 \cdot 20}

z = \frac{- ( - 31 ) + 1}{2 \cdot 20} : \frac{4}{5}

z = \frac{- ( - 31 ) - 1}{2 \cdot 20} : \frac{3}{4}

The solutions to the quadratic equation are:

z = \frac{4}{5}, z = \frac{3}{4}

4 x^{2} - 30 = 22

4 x^{2} + 4 x + 1 = 49

z^{2} + 10 z + 6 = 0

(x - 9)^{2} = 0

x^{2} + 6 x + 254 = 0

What are the solutions to the equation 18z^2-31z+12=-2z^2 ?
The solutions to the equation 18z^2-31z+12=-2z^2 are z= 4/5 ,z= 3/4
Find the zeros of 18z^2-31z+12=-2z^2
The zeros of 18z^2-31z+12=-2z^2 are z= 4/5 ,z= 3/4