What are the solutions to the equation y^2+6y-112=0 ?

The solutions to the equation y^2+6y-112=0 are y=8,y=-14

Find the zeros of y^2+6y-112=0

The zeros of y^2+6y-112=0 are y=8,y=-14

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y^{2} + 6 y - 112 = 0

y = 8, y = - 14

Solution steps

y^{2} + 6 y - 112 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 112 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 112) = 22

y_{1, 2} = \frac{- 6 \pm 22}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 6 + 22}{2 \cdot 1}, y_{2} = \frac{- 6 - 22}{2 \cdot 1}

y = \frac{- 6 + 22}{2 \cdot 1} : 8

y = \frac{- 6 - 22}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

y = 8, y = - 14

(1 - x) (3 - x) = 8

5 x^{2} - 7 x - 10 = 0

0.012 q^{2} + 50 = 60

- 3 x^{2} + 114 x + 240 = 0

r, A = 2 π r h + 2 π r^{2}

What are the solutions to the equation y^2+6y-112=0 ?
The solutions to the equation y^2+6y-112=0 are y=8,y=-14
Find the zeros of y^2+6y-112=0
The zeros of y^2+6y-112=0 are y=8,y=-14