What are the solutions to the equation 0=y^2-4y-5 ?

The solutions to the equation 0=y^2-4y-5 are y=5,y=-1

Find the zeros of 0=y^2-4y-5

The zeros of 0=y^2-4y-5 are y=5,y=-1

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0 = y^{2} - 4 y - 5

y = 5, y = - 1

Solution steps

0 = y^{2} - 4 y - 5

Switch sides

y^{2} - 4 y - 5 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 5) = 6

y_{1, 2} = \frac{- ( - 4 ) \pm 6}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 6}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 6}{2 \cdot 1}

y = \frac{- ( - 4 ) + 6}{2 \cdot 1} : 5

y = \frac{- ( - 4 ) - 6}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

y = 5, y = - 1

7^{2} = 4^{2} + b^{2}

x^{2} - 8 x - 9 = - 7

5 x^{2} + (2 (\frac{3}{2}) - 3) x + 2001 = 0

x, y = x^{2} - 3 x - 2

14500 = 4.7 \cdot 1 5^{2} + 0.22 y^{2} - 32 \cdot 15 - 1.7 y + 1550

What are the solutions to the equation 0=y^2-4y-5 ?
The solutions to the equation 0=y^2-4y-5 are y=5,y=-1
Find the zeros of 0=y^2-4y-5
The zeros of 0=y^2-4y-5 are y=5,y=-1