What are the solutions to the equation 3y^2-20y+9=-y^2+8y-40 ?

The solutions to the equation 3y^2-20y+9=-y^2+8y-40 are y= 7/2

Find the zeros of 3y^2-20y+9=-y^2+8y-40

The zeros of 3y^2-20y+9=-y^2+8y-40 are y= 7/2

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3 y^{2} - 20 y + 9 = - y^{2} + 8 y - 40

y = \frac{7}{2}

Decimal Notation

y = 3.5

Solution steps

3 y^{2} - 20 y + 9 = - y^{2} + 8 y - 40

Move

40

to the left side

3 y^{2} - 20 y + 49 = - y^{2} + 8 y

Move

8 y

to the left side

3 y^{2} - 28 y + 49 = - y^{2}

Move

y^{2}

to the left side

4 y^{2} - 28 y + 49 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 49}{2 \cdot 4}

(- 28)^{2} - 4 \cdot 4 \cdot 49 = 0

y_{1, 2} = \frac{- ( - 28 ) \pm 0}{2 \cdot 4}

y = \frac{- ( - 28 )}{2 \cdot 4}

\frac{- ( - 28 )}{2 \cdot 4} = \frac{7}{2}

y = \frac{7}{2}

The solution to the quadratic equation is:

y = \frac{7}{2}

5 u (u + 2) + 5 = 0

(a x + b)^{2} - (b x + a)^{2} = 0

28 x^{2} - 21 x = 20 x - 15

(x - 2) (x + 4) = 5 x + 41

3 x^{2} + 2 x = 9

What are the solutions to the equation 3y^2-20y+9=-y^2+8y-40 ?
The solutions to the equation 3y^2-20y+9=-y^2+8y-40 are y= 7/2
Find the zeros of 3y^2-20y+9=-y^2+8y-40
The zeros of 3y^2-20y+9=-y^2+8y-40 are y= 7/2