What are the solutions to the equation 2y^2-7y+5=0 ?

The solutions to the equation 2y^2-7y+5=0 are y= 5/2 ,y=1

Find the zeros of 2y^2-7y+5=0

The zeros of 2y^2-7y+5=0 are y= 5/2 ,y=1

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2 y^{2} - 7 y + 5 = 0

y = \frac{5}{2}, y = 1

Decimal Notation

y = 2.5, y = 1

Solution steps

2 y^{2} - 7 y + 5 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 2 \cdot 5 = 3

y_{1, 2} = \frac{- ( - 7 ) \pm 3}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 7 ) + 3}{2 \cdot 2}, y_{2} = \frac{- ( - 7 ) - 3}{2 \cdot 2}

y = \frac{- ( - 7 ) + 3}{2 \cdot 2} : \frac{5}{2}

y = \frac{- ( - 7 ) - 3}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

y = \frac{5}{2}, y = 1

25 x^{2} - 4 x = 0

x^{2} + 11 a x + 10 a^{2} = 0

x^{2} - x - 6 = - 4

0 = 5 x^{2} - 20 x + 4

3 k^{2} + k - 11 = - 4

What are the solutions to the equation 2y^2-7y+5=0 ?
The solutions to the equation 2y^2-7y+5=0 are y= 5/2 ,y=1
Find the zeros of 2y^2-7y+5=0
The zeros of 2y^2-7y+5=0 are y= 5/2 ,y=1