What are the solutions to the equation -11y+y^2-152=0 ?

The solutions to the equation -11y+y^2-152=0 are y=19,y=-8

Find the zeros of -11y+y^2-152=0

The zeros of -11y+y^2-152=0 are y=19,y=-8

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- 11 y + y^{2} - 152 = 0

y = 19, y = - 8

Solution steps

- 11 y + y^{2} - 152 = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} - 11 y - 152 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot ( - 152 )}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot (- 152) = 27

y_{1, 2} = \frac{- ( - 11 ) \pm 27}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 11 ) + 27}{2 \cdot 1}, y_{2} = \frac{- ( - 11 ) - 27}{2 \cdot 1}

y = \frac{- ( - 11 ) + 27}{2 \cdot 1} : 19

y = \frac{- ( - 11 ) - 27}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

y = 19, y = - 8

(x - 7) (x + 4)^{2} (x - 1) \geq 0

- \frac{1}{- 3}

\frac{13 !}{( 13 - 4 ) ! \cdot 4 !}

\frac{x}{3} - x = \frac{- x + 19}{5}

v^{2} = \frac{3.9388}{0.0452}

What are the solutions to the equation -11y+y^2-152=0 ?
The solutions to the equation -11y+y^2-152=0 are y=19,y=-8
Find the zeros of -11y+y^2-152=0
The zeros of -11y+y^2-152=0 are y=19,y=-8