What are the solutions to the equation y^2-12y=-19 ?

The solutions to the equation y^2-12y=-19 are y=6+sqrt(17),y=6-sqrt(17)

Find the zeros of y^2-12y=-19

The zeros of y^2-12y=-19 are y=6+sqrt(17),y=6-sqrt(17)

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y^{2} - 12 y = - 19

y = 6 + 17, y = 6 - 17

Decimal Notation

y = 10.12310 \dots, y = 1.87689 \dots

Solution steps

y^{2} - 12 y = - 19

Move

19

to the left side

y^{2} - 12 y + 19 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 19 = 217

y_{1, 2} = \frac{- ( - 12 ) \pm 2 17}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 12 ) + 2 17}{2 \cdot 1}, y_{2} = \frac{- ( - 12 ) - 2 17}{2 \cdot 1}

y = \frac{- ( - 12 ) + 2 17}{2 \cdot 1} : 6 + 17

y = \frac{- ( - 12 ) - 2 17}{2 \cdot 1} : 6 - 17

The solutions to the quadratic equation are:

y = 6 + 17, y = 6 - 17

11

\frac{3 x - 1}{x + 2} \geq 1

x^{2} + 31 = 0

3 x + 13 = 2 (x + 9)

h^{2} - 30 h - 120 = 0

What are the solutions to the equation y^2-12y=-19 ?
The solutions to the equation y^2-12y=-19 are y=6+sqrt(17),y=6-sqrt(17)
Find the zeros of y^2-12y=-19
The zeros of y^2-12y=-19 are y=6+sqrt(17),y=6-sqrt(17)