What are the solutions to the equation y^2+28y-27=0 ?

The solutions to the equation y^2+28y-27=0 are y=-14+sqrt(223),y=-14-sqrt(223)

Find the zeros of y^2+28y-27=0

The zeros of y^2+28y-27=0 are y=-14+sqrt(223),y=-14-sqrt(223)

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y^{2} + 28 y - 27 = 0

y = - 14 + 223, y = - 14 - 223

Solution steps

y^{2} + 28 y - 27 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 1 \cdot ( - 27 )}{2 \cdot 1}

2 8^{2} - 4 \cdot 1 \cdot (- 27) = 2223

y_{1, 2} = \frac{- 28 \pm 2 223}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 28 + 2 223}{2 \cdot 1}, y_{2} = \frac{- 28 - 2 223}{2 \cdot 1}

y = \frac{- 28 + 2 223}{2 \cdot 1} : - 14 + 223

y = \frac{- 28 - 2 223}{2 \cdot 1} : - 14 - 223

The solutions to the quadratic equation are:

y = - 14 + 223, y = - 14 - 223

x + 7 > 15 and - 3 x + 4 \geq 7

2^{1.5}

3 x^{2} - 5 x = 12

∣ x - 2 ∣ - 2 = 4

- 1 \leq \frac{4 x - 6}{- 3} \leq 0

What are the solutions to the equation y^2+28y-27=0 ?
The solutions to the equation y^2+28y-27=0 are y=-14+sqrt(223),y=-14-sqrt(223)
Find the zeros of y^2+28y-27=0
The zeros of y^2+28y-27=0 are y=-14+sqrt(223),y=-14-sqrt(223)