What are the solutions to the equation y^2+8(y-1)=3 ?

The solutions to the equation y^2+8(y-1)=3 are y=-4+3sqrt(3),y=-4-3sqrt(3)

Find the zeros of y^2+8(y-1)=3

The zeros of y^2+8(y-1)=3 are y=-4+3sqrt(3),y=-4-3sqrt(3)

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y^{2} + 8 (y - 1) = 3

y = - 4 + 33, y = - 4 - 33

Decimal Notation

y = 1.19615 \dots, y = - 9.19615 \dots

Solution steps

y^{2} + 8 (y - 1) = 3

Expand

y^{2} + 8 (y - 1) : y^{2} + 8 y - 8

y^{2} + 8 y - 8 = 3

Move

3

to the left side

y^{2} + 8 y - 11 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 11) = 63

y_{1, 2} = \frac{- 8 \pm 6 3}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 8 + 6 3}{2 \cdot 1}, y_{2} = \frac{- 8 - 6 3}{2 \cdot 1}

y = \frac{- 8 + 6 3}{2 \cdot 1} : - 4 + 33

y = \frac{- 8 - 6 3}{2 \cdot 1} : - 4 - 33

The solutions to the quadratic equation are:

y = - 4 + 33, y = - 4 - 33

2^{- 1}

- 2 t^{2} = - 18

5 m^{2} + 12 m + 7

\frac{x + 2}{24} = \frac{9}{2}

15 < ∣ - 2 d - 3 ∣ + 6

What are the solutions to the equation y^2+8(y-1)=3 ?
The solutions to the equation y^2+8(y-1)=3 are y=-4+3sqrt(3),y=-4-3sqrt(3)
Find the zeros of y^2+8(y-1)=3
The zeros of y^2+8(y-1)=3 are y=-4+3sqrt(3),y=-4-3sqrt(3)