What are the solutions to the equation 6y^2+7y-3=0 ?

The solutions to the equation 6y^2+7y-3=0 are y= 1/3 ,y=-3/2

Find the zeros of 6y^2+7y-3=0

The zeros of 6y^2+7y-3=0 are y= 1/3 ,y=-3/2

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6 y^{2} + 7 y - 3 = 0

y = \frac{1}{3}, y = - \frac{3}{2}

Solution steps

6 y^{2} + 7 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

7^{2} - 4 \cdot 6 (- 3) = 11

y_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- 7 + 11}{2 \cdot 6}, y_{2} = \frac{- 7 - 11}{2 \cdot 6}

y = \frac{- 7 + 11}{2 \cdot 6} : \frac{1}{3}

y = \frac{- 7 - 11}{2 \cdot 6} : - \frac{3}{2}

The solutions to the quadratic equation are:

y = \frac{1}{3}, y = - \frac{3}{2}

(\frac{2}{9})^{2} + \frac{3}{4} - (\frac{1}{3})^{2}

∣ 5 a + 5 ∣ = ∣ 6 a - 7 ∣

(2 x + 1)^{2} = 81

(2 x - 4)^{2} - 10 (2 x - 4) + 21 = 0

2 x^{2} = 3 x

What are the solutions to the equation 6y^2+7y-3=0 ?
The solutions to the equation 6y^2+7y-3=0 are y= 1/3 ,y=-3/2
Find the zeros of 6y^2+7y-3=0
The zeros of 6y^2+7y-3=0 are y= 1/3 ,y=-3/2