What are the solutions to the equation y^2-14y=-38 ?

The solutions to the equation y^2-14y=-38 are y=7+sqrt(11),y=7-sqrt(11)

Find the zeros of y^2-14y=-38

The zeros of y^2-14y=-38 are y=7+sqrt(11),y=7-sqrt(11)

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y^{2} - 14 y = - 38

y = 7 + 11, y = 7 - 11

Decimal Notation

y = 10.31662 \dots, y = 3.68337 \dots

Solution steps

y^{2} - 14 y = - 38

Move

38

to the left side

y^{2} - 14 y + 38 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 38}{2 \cdot 1}

(- 14)^{2} - 4 \cdot 1 \cdot 38 = 211

y_{1, 2} = \frac{- ( - 14 ) \pm 2 11}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 14 ) + 2 11}{2 \cdot 1}, y_{2} = \frac{- ( - 14 ) - 2 11}{2 \cdot 1}

y = \frac{- ( - 14 ) + 2 11}{2 \cdot 1} : 7 + 11

y = \frac{- ( - 14 ) - 2 11}{2 \cdot 1} : 7 - 11

The solutions to the quadratic equation are:

y = 7 + 11, y = 7 - 11

∣ 5 u - 15 ∣ = 10

\frac{3 8}{8 5}

\frac{x - 1}{x + 2} \leq 0

(- \frac{4}{3} x y^{3}) \cdot (9 x^{8} y)

(4 w) (10 w - 7) = 0

What are the solutions to the equation y^2-14y=-38 ?
The solutions to the equation y^2-14y=-38 are y=7+sqrt(11),y=7-sqrt(11)
Find the zeros of y^2-14y=-38
The zeros of y^2-14y=-38 are y=7+sqrt(11),y=7-sqrt(11)