What are the solutions to the equation 5w^2-4w+1=3w ?

The solutions to the equation 5w^2-4w+1=3w are w=(7+sqrt(29))/(10),w=(7-sqrt(29))/(10)

Find the zeros of 5w^2-4w+1=3w

The zeros of 5w^2-4w+1=3w are w=(7+sqrt(29))/(10),w=(7-sqrt(29))/(10)

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5 w^{2} - 4 w + 1 = 3 w

w = \frac{7 + 29}{10}, w = \frac{7 - 29}{10}

Decimal Notation

w = 1.23851 \dots, w = 0.16148 \dots

Solution steps

5 w^{2} - 4 w + 1 = 3 w

Move

3 w

to the left side

5 w^{2} - 7 w + 1 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 \cdot 1 = 29

w_{1, 2} = \frac{- ( - 7 ) \pm 29}{2 \cdot 5}

Separate the solutions

w_{1} = \frac{- ( - 7 ) + 29}{2 \cdot 5}, w_{2} = \frac{- ( - 7 ) - 29}{2 \cdot 5}

w = \frac{- ( - 7 ) + 29}{2 \cdot 5} : \frac{7 + 29}{10}

w = \frac{- ( - 7 ) - 29}{2 \cdot 5} : \frac{7 - 29}{10}

The solutions to the quadratic equation are:

w = \frac{7 + 29}{10}, w = \frac{7 - 29}{10}

- x^{3} + 9 x^{2} + 10 x

r - 7 = 2

- 3 \cdot 2

x^{2} - 6 x + 9 > 0

∣ 2 - 3 x ∣ \geq 5

What are the solutions to the equation 5w^2-4w+1=3w ?
The solutions to the equation 5w^2-4w+1=3w are w=(7+sqrt(29))/(10),w=(7-sqrt(29))/(10)
Find the zeros of 5w^2-4w+1=3w
The zeros of 5w^2-4w+1=3w are w=(7+sqrt(29))/(10),w=(7-sqrt(29))/(10)