What are the solutions to the equation 6y+10=y^2+3y ?

The solutions to the equation 6y+10=y^2+3y are y=5,y=-2

Find the zeros of 6y+10=y^2+3y

The zeros of 6y+10=y^2+3y are y=5,y=-2

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6 y + 10 = y^{2} + 3 y

y = 5, y = - 2

Solution steps

6 y + 10 = y^{2} + 3 y

Switch sides

y^{2} + 3 y = 6 y + 10

Move

10

to the left side

y^{2} + 3 y - 10 = 6 y

Move

6 y

to the left side

y^{2} - 3 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 10) = 7

y_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 1}

y = \frac{- ( - 3 ) + 7}{2 \cdot 1} : 5

y = \frac{- ( - 3 ) - 7}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

y = 5, y = - 2

16 y^{2} - 25 = 0

(4 x + 9)^{2} = 25

(x + 10) (x - 5) = 0

x^{2} + 5 x = 126

x^{2} - 4 x + 27 = 0

What are the solutions to the equation 6y+10=y^2+3y ?
The solutions to the equation 6y+10=y^2+3y are y=5,y=-2
Find the zeros of 6y+10=y^2+3y
The zeros of 6y+10=y^2+3y are y=5,y=-2