What are the solutions to the equation y^2+13y=-42 ?

The solutions to the equation y^2+13y=-42 are y=-6,y=-7

Find the zeros of y^2+13y=-42

The zeros of y^2+13y=-42 are y=-6,y=-7

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y^{2} + 13 y = - 42

y = - 6, y = - 7

Solution steps

y^{2} + 13 y = - 42

Move

42

to the left side

y^{2} + 13 y + 42 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 42 = 1

y_{1, 2} = \frac{- 13 \pm 1}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 13 + 1}{2 \cdot 1}, y_{2} = \frac{- 13 - 1}{2 \cdot 1}

y = \frac{- 13 + 1}{2 \cdot 1} : - 6

y = \frac{- 13 - 1}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

y = - 6, y = - 7

5 ∣ 2 x - 1 ∣ = 15

x^{2} - 14 x + 33 = 0

\frac{29}{4} \cdot 2

2 (m - 1) - 3 (m - 1)^{2} + 2 (m - 1)^{3}

lo g_{2} (5) = x

What are the solutions to the equation y^2+13y=-42 ?
The solutions to the equation y^2+13y=-42 are y=-6,y=-7
Find the zeros of y^2+13y=-42
The zeros of y^2+13y=-42 are y=-6,y=-7