What are the solutions to the equation y^2+y=20 ?

The solutions to the equation y^2+y=20 are y=4,y=-5

Find the zeros of y^2+y=20

The zeros of y^2+y=20 are y=4,y=-5

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y^{2} + y = 20

y = 4, y = - 5

Solution steps

y^{2} + y = 20

Move

20

to the left side

y^{2} + y - 20 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 20 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 20) = 9

y_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 9}{2 \cdot 1}, y_{2} = \frac{- 1 - 9}{2 \cdot 1}

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

y = \frac{- 1 - 9}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

y = 4, y = - 5

4 a^{2} - 25 b^{2}

\frac{8}{5} \div (\frac{3}{4} \div \frac{3}{8})

\frac{2 b - 3}{7} - \frac{b}{2} = \frac{b + 3}{14}

618

15 + 6 b - 6 b = 4 b - 1 - 6 b

What are the solutions to the equation y^2+y=20 ?
The solutions to the equation y^2+y=20 are y=4,y=-5
Find the zeros of y^2+y=20
The zeros of y^2+y=20 are y=4,y=-5