What are the solutions to the equation 25z^2+222=150z ?

The solutions to the equation 25z^2+222=150z are z=(15+sqrt(3))/5 ,z=(15-sqrt(3))/5

Find the zeros of 25z^2+222=150z

The zeros of 25z^2+222=150z are z=(15+sqrt(3))/5 ,z=(15-sqrt(3))/5

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25 z^{2} + 222 = 150 z

z = \frac{15 + 3}{5}, z = \frac{15 - 3}{5}

Decimal Notation

z = 3.34641 \dots, z = 2.65358 \dots

Solution steps

25 z^{2} + 222 = 150 z

Move

150 z

to the left side

25 z^{2} + 222 - 150 z = 0

Write in the standard form

a x^{2} + b x + c = 0

25 z^{2} - 150 z + 222 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 150 ) \pm ( - 150 ) ^{2} - 4 \cdot 25 \cdot 222}{2 \cdot 25}

(- 150)^{2} - 4 \cdot 25 \cdot 222 = 103

z_{1, 2} = \frac{- ( - 150 ) \pm 10 3}{2 \cdot 25}

Separate the solutions

z_{1} = \frac{- ( - 150 ) + 10 3}{2 \cdot 25}, z_{2} = \frac{- ( - 150 ) - 10 3}{2 \cdot 25}

z = \frac{- ( - 150 ) + 10 3}{2 \cdot 25} : \frac{15 + 3}{5}

z = \frac{- ( - 150 ) - 10 3}{2 \cdot 25} : \frac{15 - 3}{5}

The solutions to the quadratic equation are:

z = \frac{15 + 3}{5}, z = \frac{15 - 3}{5}

\frac{- 6 ( 0.74522293 ) + 30}{12}

3 x + 3 = x + 13

∣ x - 6 ∣ \leq 4

(4 \frac{x}{2}) (4 \frac{x}{5}) = 4^{8}

\frac{1}{4} - \frac{1}{5} + \frac{1}{6} - \frac{1}{8}

What are the solutions to the equation 25z^2+222=150z ?
The solutions to the equation 25z^2+222=150z are z=(15+sqrt(3))/5 ,z=(15-sqrt(3))/5
Find the zeros of 25z^2+222=150z
The zeros of 25z^2+222=150z are z=(15+sqrt(3))/5 ,z=(15-sqrt(3))/5