What are the solutions to the equation 15y^2-30=22y+7 ?

The solutions to the equation 15y^2-30=22y+7 are y= 37/15 ,y=-1

Find the zeros of 15y^2-30=22y+7

The zeros of 15y^2-30=22y+7 are y= 37/15 ,y=-1

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15 y^{2} - 30 = 22 y + 7

y = \frac{37}{15}, y = - 1

Solution steps

15 y^{2} - 30 = 22 y + 7

Move

7

to the left side

15 y^{2} - 37 = 22 y

Move

22 y

to the left side

15 y^{2} - 37 - 22 y = 0

Write in the standard form

a x^{2} + b x + c = 0

15 y^{2} - 22 y - 37 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 15 ( - 37 )}{2 \cdot 15}

(- 22)^{2} - 4 \cdot 15 (- 37) = 52

y_{1, 2} = \frac{- ( - 22 ) \pm 52}{2 \cdot 15}

Separate the solutions

y_{1} = \frac{- ( - 22 ) + 52}{2 \cdot 15}, y_{2} = \frac{- ( - 22 ) - 52}{2 \cdot 15}

y = \frac{- ( - 22 ) + 52}{2 \cdot 15} : \frac{37}{15}

y = \frac{- ( - 22 ) - 52}{2 \cdot 15} : - 1

The solutions to the quadratic equation are:

y = \frac{37}{15}, y = - 1

3 x + 23 = 2 x + 59

x = x + \frac{x}{3} + 1

3 (x - 4) = 8 - 5 (3 x + 1)

6 - 3 ∣ 2 x + 6 ∣ = 0

6128

What are the solutions to the equation 15y^2-30=22y+7 ?
The solutions to the equation 15y^2-30=22y+7 are y= 37/15 ,y=-1
Find the zeros of 15y^2-30=22y+7
The zeros of 15y^2-30=22y+7 are y= 37/15 ,y=-1