What are the solutions to the equation -y^2+25y-144=0 ?

The solutions to the equation -y^2+25y-144=0 are y=9,y=16

Find the zeros of -y^2+25y-144=0

The zeros of -y^2+25y-144=0 are y=9,y=16

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- y^{2} + 25 y - 144 = 0

y = 9, y = 16

Solution steps

- y^{2} + 25 y - 144 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 ( - 1 ) ( - 144 )}{2 ( - 1 )}

2 5^{2} - 4 (- 1) (- 144) = 7

y_{1, 2} = \frac{- 25 \pm 7}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 25 + 7}{2 ( - 1 )}, y_{2} = \frac{- 25 - 7}{2 ( - 1 )}

y = \frac{- 25 + 7}{2 ( - 1 )} : 9

y = \frac{- 25 - 7}{2 ( - 1 )} : 16

The solutions to the quadratic equation are:

y = 9, y = 16

x^{2} + x + 9 = 20

4 z^{2} + 2 z + 1 = 0

n^{2} - 5 n = - 4

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

x^{2} + 14 x + 21 = 0

What are the solutions to the equation -y^2+25y-144=0 ?
The solutions to the equation -y^2+25y-144=0 are y=9,y=16
Find the zeros of -y^2+25y-144=0
The zeros of -y^2+25y-144=0 are y=9,y=16