What are the solutions to the equation b=(y+16)(y+24)+16 ?

The solutions to the equation b=(y+16)(y+24)+16 are y=-20+sqrt(b),y=-sqrt(b)-20

Find the zeros of b=(y+16)(y+24)+16

The zeros of b=(y+16)(y+24)+16 are y=-20+sqrt(b),y=-sqrt(b)-20

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b = (y + 16) (y + 24) + 16

y = - 20 + b, y = - b - 20

Solution steps

b = (y + 16) (y + 24) + 16

Expand

(y + 16) (y + 24) + 16 : y^{2} + 40 y + 400

b = y^{2} + 40 y + 400

Switch sides

y^{2} + 40 y + 400 = b

Move

b

to the left side

y^{2} + 40 y + 400 - b = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 1 \cdot ( 400 - b )}{2 \cdot 1}

Simplify

4 0^{2} - 4 \cdot 1 \cdot (400 - b) : 2 b

y_{1, 2} = \frac{- 40 \pm 2 b}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 40 + 2 b}{2 \cdot 1}, y_{2} = \frac{- 40 - 2 b}{2 \cdot 1}

y = \frac{- 40 + 2 b}{2 \cdot 1} : - 20 + b

y = \frac{- 40 - 2 b}{2 \cdot 1} : - b - 20

The solutions to the quadratic equation are:

y = - 20 + b, y = - b - 20

x, f = x^{2} - p - c

2 p^{2} + 5 p - 2 = 0

21 n = 192 - x^{2}

x (\frac{3 x + 1}{2}) = 6

a = x^{2} - 4 x + 1

What are the solutions to the equation b=(y+16)(y+24)+16 ?
The solutions to the equation b=(y+16)(y+24)+16 are y=-20+sqrt(b),y=-sqrt(b)-20
Find the zeros of b=(y+16)(y+24)+16
The zeros of b=(y+16)(y+24)+16 are y=-20+sqrt(b),y=-sqrt(b)-20