What are the solutions to the equation 5z^2-21z+11=-4z ?

The solutions to the equation 5z^2-21z+11=-4z are z=(17+sqrt(69))/(10),z=(17-sqrt(69))/(10)

Find the zeros of 5z^2-21z+11=-4z

The zeros of 5z^2-21z+11=-4z are z=(17+sqrt(69))/(10),z=(17-sqrt(69))/(10)

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5 z^{2} - 21 z + 11 = - 4 z

z = \frac{17 + 69}{10}, z = \frac{17 - 69}{10}

Decimal Notation

z = 2.53066 \dots, z = 0.86933 \dots

Solution steps

5 z^{2} - 21 z + 11 = - 4 z

Move

4 z

to the left side

5 z^{2} - 17 z + 11 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 5 \cdot 11}{2 \cdot 5}

(- 17)^{2} - 4 \cdot 5 \cdot 11 = 69

z_{1, 2} = \frac{- ( - 17 ) \pm 69}{2 \cdot 5}

Separate the solutions

z_{1} = \frac{- ( - 17 ) + 69}{2 \cdot 5}, z_{2} = \frac{- ( - 17 ) - 69}{2 \cdot 5}

z = \frac{- ( - 17 ) + 69}{2 \cdot 5} : \frac{17 + 69}{10}

z = \frac{- ( - 17 ) - 69}{2 \cdot 5} : \frac{17 - 69}{10}

The solutions to the quadratic equation are:

z = \frac{17 + 69}{10}, z = \frac{17 - 69}{10}

2 x^{2} + 11 x - 63 = 0

a^{2} + 4 = 4 a

3 x^{2} - 12 x = - 8

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

(x + 2)^{2} - 5 = 31

What are the solutions to the equation 5z^2-21z+11=-4z ?
The solutions to the equation 5z^2-21z+11=-4z are z=(17+sqrt(69))/(10),z=(17-sqrt(69))/(10)
Find the zeros of 5z^2-21z+11=-4z
The zeros of 5z^2-21z+11=-4z are z=(17+sqrt(69))/(10),z=(17-sqrt(69))/(10)