What are the solutions to the equation y^2+8y=10 ?

The solutions to the equation y^2+8y=10 are y=-4+sqrt(26),y=-4-sqrt(26)

Find the zeros of y^2+8y=10

The zeros of y^2+8y=10 are y=-4+sqrt(26),y=-4-sqrt(26)

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y^{2} + 8 y = 10

y = - 4 + 26, y = - 4 - 26

Decimal Notation

y = 1.09901 \dots, y = - 9.09901 \dots

Solution steps

y^{2} + 8 y = 10

Move

10

to the left side

y^{2} + 8 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 10) = 226

y_{1, 2} = \frac{- 8 \pm 2 26}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 8 + 2 26}{2 \cdot 1}, y_{2} = \frac{- 8 - 2 26}{2 \cdot 1}

y = \frac{- 8 + 2 26}{2 \cdot 1} : - 4 + 26

y = \frac{- 8 - 2 26}{2 \cdot 1} : - 4 - 26

The solutions to the quadratic equation are:

y = - 4 + 26, y = - 4 - 26

5 n^{2} - 20

- 6 x - 12 < 54

4 r^{2} + 3 r - 9 = 0

x (x - 2) - x = 0

3 x - 6 = 84 - 7 x

What are the solutions to the equation y^2+8y=10 ?
The solutions to the equation y^2+8y=10 are y=-4+sqrt(26),y=-4-sqrt(26)
Find the zeros of y^2+8y=10
The zeros of y^2+8y=10 are y=-4+sqrt(26),y=-4-sqrt(26)