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

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

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

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

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y^{2} - 4 y = 5

y = 5, y = - 1

Solution steps

y^{2} - 4 y = 5

Move

5

to the left side

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

x^{2} - 18 (x - 4) = 0

3 a^{2} - 8 a + 4 = 0

x^{2} + 2 x + 5 = - 5

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

63 (p^{2} - 1) = 32 p

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