What are the solutions to the equation y^2-4y-45=0 ?

The solutions to the equation y^2-4y-45=0 are y=9,y=-5

Find the zeros of y^2-4y-45=0

The zeros of y^2-4y-45=0 are y=9,y=-5

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y^{2} - 4 y - 45 = 0

y = 9, y = - 5

Solution steps

y^{2} - 4 y - 45 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 45 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 45) = 14

y_{1, 2} = \frac{- ( - 4 ) \pm 14}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 14}{2 \cdot 1}, y_{2} = \frac{- ( - 4 ) - 14}{2 \cdot 1}

y = \frac{- ( - 4 ) + 14}{2 \cdot 1} : 9

y = \frac{- ( - 4 ) - 14}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

y = 9, y = - 5

3 x^{2} + 9 x + 7 = 0

3 x^{2} = 27 x

\frac{2}{3} x^{2} + 20 = 2

6 x^{2} - 1 x - 2 = 0

(D + \frac{2}{25}) x^{1} - \frac{1}{50} x^{2} = 0

What are the solutions to the equation y^2-4y-45=0 ?
The solutions to the equation y^2-4y-45=0 are y=9,y=-5
Find the zeros of y^2-4y-45=0
The zeros of y^2-4y-45=0 are y=9,y=-5