What are the solutions to the equation -5z^2-4z+3=0 ?

The solutions to the equation -5z^2-4z+3=0 are z=-(2+sqrt(19))/5 ,z=(sqrt(19)-2)/5

Find the zeros of -5z^2-4z+3=0

The zeros of -5z^2-4z+3=0 are z=-(2+sqrt(19))/5 ,z=(sqrt(19)-2)/5

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

z = - \frac{2 + 19}{5}, z = \frac{19 - 2}{5}

Decimal Notation

z = - 1.27177 \dots, z = 0.47177 \dots

Solution steps

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

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 5 ) \cdot 3}{2 ( - 5 )}

(- 4)^{2} - 4 (- 5) \cdot 3 = 219

z_{1, 2} = \frac{- ( - 4 ) \pm 2 19}{2 ( - 5 )}

Separate the solutions

z_{1} = \frac{- ( - 4 ) + 2 19}{2 ( - 5 )}, z_{2} = \frac{- ( - 4 ) - 2 19}{2 ( - 5 )}

z = \frac{- ( - 4 ) + 2 19}{2 ( - 5 )} : - \frac{2 + 19}{5}

z = \frac{- ( - 4 ) - 2 19}{2 ( - 5 )} : \frac{19 - 2}{5}

The solutions to the quadratic equation are:

z = - \frac{2 + 19}{5}, z = \frac{19 - 2}{5}

- 5 x^{2} = 125

x^{2} + 16 x + 64 = 2 x^{2} - 32 + 256

y, x = 1 - (7 y - 4)^{2}

x, 15 x^{2} y + 22 x y - 6 = 0

- x^{(2)} + 4 x = 0

What are the solutions to the equation -5z^2-4z+3=0 ?
The solutions to the equation -5z^2-4z+3=0 are z=-(2+sqrt(19))/5 ,z=(sqrt(19)-2)/5
Find the zeros of -5z^2-4z+3=0
The zeros of -5z^2-4z+3=0 are z=-(2+sqrt(19))/5 ,z=(sqrt(19)-2)/5