What are the solutions to the equation y^2+10y-26=0 ?

The solutions to the equation y^2+10y-26=0 are y=-5+sqrt(51),y=-5-sqrt(51)

Find the zeros of y^2+10y-26=0

The zeros of y^2+10y-26=0 are y=-5+sqrt(51),y=-5-sqrt(51)

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y^{2} + 10 y - 26 = 0

y = - 5 + 51, y = - 5 - 51

Decimal Notation

y = 2.14142 \dots, y = - 12.14142 \dots

Solution steps

y^{2} + 10 y - 26 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 26 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 26) = 251

y_{1, 2} = \frac{- 10 \pm 2 51}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 10 + 2 51}{2 \cdot 1}, y_{2} = \frac{- 10 - 2 51}{2 \cdot 1}

y = \frac{- 10 + 2 51}{2 \cdot 1} : - 5 + 51

y = \frac{- 10 - 2 51}{2 \cdot 1} : - 5 - 51

The solutions to the quadratic equation are:

y = - 5 + 51, y = - 5 - 51

- 3 x^{2} - 51 x - 216 = 0

6 (x + 12) = x^{2}

m^{2} + 6 m + 18 = 0

x^{2} + 7 x + \frac{49}{4} = 4

(x + 5)^{2} = 10 x + 50

What are the solutions to the equation y^2+10y-26=0 ?
The solutions to the equation y^2+10y-26=0 are y=-5+sqrt(51),y=-5-sqrt(51)
Find the zeros of y^2+10y-26=0
The zeros of y^2+10y-26=0 are y=-5+sqrt(51),y=-5-sqrt(51)