What are the solutions to the equation y^2-5y=3 ?

The solutions to the equation y^2-5y=3 are y=(5+sqrt(37))/2 ,y=(5-sqrt(37))/2

Find the zeros of y^2-5y=3

The zeros of y^2-5y=3 are y=(5+sqrt(37))/2 ,y=(5-sqrt(37))/2

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y^{2} - 5 y = 3

y = \frac{5 + 37}{2}, y = \frac{5 - 37}{2}

Decimal Notation

y = 5.54138 \dots, y = - 0.54138 \dots

Solution steps

y^{2} - 5 y = 3

Move

3

to the left side

y^{2} - 5 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 3) = 37

y_{1, 2} = \frac{- ( - 5 ) \pm 37}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 37}{2 \cdot 1}, y_{2} = \frac{- ( - 5 ) - 37}{2 \cdot 1}

y = \frac{- ( - 5 ) + 37}{2 \cdot 1} : \frac{5 + 37}{2}

y = \frac{- ( - 5 ) - 37}{2 \cdot 1} : \frac{5 - 37}{2}

The solutions to the quadratic equation are:

y = \frac{5 + 37}{2}, y = \frac{5 - 37}{2}

m^{2} - 17 m + 72

9 \geq 9

x^{2} - 2 x = 3 x - 6

\frac{4 y z ^{2}}{3 x y ^{4} z} \cdot \frac{15 x ^{2} y}{8 x ^{3} yz}

3 \geq 2 x

What are the solutions to the equation y^2-5y=3 ?
The solutions to the equation y^2-5y=3 are y=(5+sqrt(37))/2 ,y=(5-sqrt(37))/2
Find the zeros of y^2-5y=3
The zeros of y^2-5y=3 are y=(5+sqrt(37))/2 ,y=(5-sqrt(37))/2