What are the solutions to the equation 4y+3y^2=15 ?

The solutions to the equation 4y+3y^2=15 are y= 5/3 ,y=-3

Find the zeros of 4y+3y^2=15

The zeros of 4y+3y^2=15 are y= 5/3 ,y=-3

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4 y + 3 y^{2} = 15

y = \frac{5}{3}, y = - 3

Solution steps

4 y + 3 y^{2} = 15

Move

15

to the left side

4 y + 3 y^{2} - 15 = 0

Write in the standard form

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

3 y^{2} + 4 y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 ( - 15 )}{2 \cdot 3}

4^{2} - 4 \cdot 3 (- 15) = 14

y_{1, 2} = \frac{- 4 \pm 14}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 4 + 14}{2 \cdot 3}, y_{2} = \frac{- 4 - 14}{2 \cdot 3}

y = \frac{- 4 + 14}{2 \cdot 3} : \frac{5}{3}

y = \frac{- 4 - 14}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

y = \frac{5}{3}, y = - 3

3 k - (6 + k) \geq 3

0. 2^{6}

\frac{3}{4} - 3

∣ 3 v + 9 ∣ = 3

2 x + 9 x - 2 = 3 x - 5 - 4

What are the solutions to the equation 4y+3y^2=15 ?
The solutions to the equation 4y+3y^2=15 are y= 5/3 ,y=-3
Find the zeros of 4y+3y^2=15
The zeros of 4y+3y^2=15 are y= 5/3 ,y=-3