What are the solutions to the equation 1.4z^2+3.1z=1 ?

The solutions to the equation 1.4z^2+3.1z=1 are z= 2/7 ,z=-5/2

Find the zeros of 1.4z^2+3.1z=1

The zeros of 1.4z^2+3.1z=1 are z= 2/7 ,z=-5/2

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1.4 z^{2} + 3.1 z = 1

z = \frac{2}{7}, z = - \frac{5}{2}

Decimal Notation

z = 0.28571 \dots, z = - 2.5

Solution steps

1.4 z^{2} + 3.1 z = 1

Multiply both sides by

10

14 z^{2} + 31 z = 10

Move

10

to the left side

14 z^{2} + 31 z - 10 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 14 ( - 10 )}{2 \cdot 14}

3 1^{2} - 4 \cdot 14 (- 10) = 39

z_{1, 2} = \frac{- 31 \pm 39}{2 \cdot 14}

Separate the solutions

z_{1} = \frac{- 31 + 39}{2 \cdot 14}, z_{2} = \frac{- 31 - 39}{2 \cdot 14}

z = \frac{- 31 + 39}{2 \cdot 14} : \frac{2}{7}

z = \frac{- 31 - 39}{2 \cdot 14} : - \frac{5}{2}

The solutions to the quadratic equation are:

z = \frac{2}{7}, z = - \frac{5}{2}

\frac{x ^{2}}{4} - x + \frac{4}{10} = 8

x^{2} - 33 x + 6 = 0

x, x^{2} - 3 x k - 1 = 0

x, 2 x^{2} + 3 x y - 2 y^{2} = 7

x, \frac{( f \cdot f ( x ^{2} ) )}{( 1 + [ f ] ^{2} )} = \frac{1}{2}

What are the solutions to the equation 1.4z^2+3.1z=1 ?
The solutions to the equation 1.4z^2+3.1z=1 are z= 2/7 ,z=-5/2
Find the zeros of 1.4z^2+3.1z=1
The zeros of 1.4z^2+3.1z=1 are z= 2/7 ,z=-5/2