What are the solutions to the equation 7w^2+12w+10=4w^2 ?

The solutions to the equation 7w^2+12w+10=4w^2 are w=(-6+sqrt(6))/3 ,w=-(2sqrt(3)+sqrt(2))/(sqrt(3))

Find the zeros of 7w^2+12w+10=4w^2

The zeros of 7w^2+12w+10=4w^2 are w=(-6+sqrt(6))/3 ,w=-(2sqrt(3)+sqrt(2))/(sqrt(3))

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7 w^{2} + 12 w + 10 = 4 w^{2}

w = \frac{- 6 + 6}{3}, w = - \frac{2 3 + 2}{3}

Decimal Notation

w = - 1.18350 \dots, w = - 2.81649 \dots

Solution steps

7 w^{2} + 12 w + 10 = 4 w^{2}

Move

4 w^{2}

to the left side

3 w^{2} + 12 w + 10 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 \cdot 10}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 \cdot 10 = 26

w_{1, 2} = \frac{- 12 \pm 2 6}{2 \cdot 3}

Separate the solutions

w_{1} = \frac{- 12 + 2 6}{2 \cdot 3}, w_{2} = \frac{- 12 - 2 6}{2 \cdot 3}

w = \frac{- 12 + 2 6}{2 \cdot 3} : \frac{- 6 + 6}{3}

w = \frac{- 12 - 2 6}{2 \cdot 3} : - \frac{2 3 + 2}{3}

The solutions to the quadratic equation are:

w = \frac{- 6 + 6}{3}, w = - \frac{2 3 + 2}{3}

2 a^{2} = 18

2 x^{2} + 10 x + 2 = 0

z^{2} - 3 z + 2 = 0

x, x^{2} + y^{2} + z^{2} = 3

x (2 x + 3) = 14

What are the solutions to the equation 7w^2+12w+10=4w^2 ?
The solutions to the equation 7w^2+12w+10=4w^2 are w=(-6+sqrt(6))/3 ,w=-(2sqrt(3)+sqrt(2))/(sqrt(3))
Find the zeros of 7w^2+12w+10=4w^2
The zeros of 7w^2+12w+10=4w^2 are w=(-6+sqrt(6))/3 ,w=-(2sqrt(3)+sqrt(2))/(sqrt(3))