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

The solutions to the equation y^2-5y=7 are y=(5+sqrt(53))/2 ,y=(5-sqrt(53))/2

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

The zeros of y^2-5y=7 are y=(5+sqrt(53))/2 ,y=(5-sqrt(53))/2

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y^{2} - 5 y = 7

y = \frac{5 + 53}{2}, y = \frac{5 - 53}{2}

Decimal Notation

y = 6.14005 \dots, y = - 1.14005 \dots

Solution steps

y^{2} - 5 y = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 1 \cdot (- 7) = 53

y_{1, 2} = \frac{- ( - 5 ) \pm 53}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 53}{2 \cdot 1}, y_{2} = \frac{- ( - 5 ) - 53}{2 \cdot 1}

y = \frac{- ( - 5 ) + 53}{2 \cdot 1} : \frac{5 + 53}{2}

y = \frac{- ( - 5 ) - 53}{2 \cdot 1} : \frac{5 - 53}{2}

The solutions to the quadratic equation are:

y = \frac{5 + 53}{2}, y = \frac{5 - 53}{2}

x, (1 - t^{2}) x^{2} - 2 t^{2} x - (t^{2} + 1) = 0

x^{2} + 100 x + 250 = 0

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

- 2 x = - x^{2}

(4 x - 1) (x + 2) = 4 x

What are the solutions to the equation y^2-5y=7 ?
The solutions to the equation y^2-5y=7 are y=(5+sqrt(53))/2 ,y=(5-sqrt(53))/2
Find the zeros of y^2-5y=7
The zeros of y^2-5y=7 are y=(5+sqrt(53))/2 ,y=(5-sqrt(53))/2