What are the solutions to the equation 3x^2+13x-28=2x^2+2 ?

The solutions to the equation 3x^2+13x-28=2x^2+2 are x=2,x=-15

Find the zeros of 3x^2+13x-28=2x^2+2

The zeros of 3x^2+13x-28=2x^2+2 are x=2,x=-15

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3 x^{2} + 13 x - 28 = 2 x^{2} + 2

x = 2, x = - 15

Solution steps

3 x^{2} + 13 x - 28 = 2 x^{2} + 2

Move

2

to the left side

3 x^{2} + 13 x - 30 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} + 13 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot (- 30) = 17

x_{1, 2} = \frac{- 13 \pm 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 17}{2 \cdot 1}, x_{2} = \frac{- 13 - 17}{2 \cdot 1}

x = \frac{- 13 + 17}{2 \cdot 1} : 2

x = \frac{- 13 - 17}{2 \cdot 1} : - 15

The solutions to the quadratic equation are:

x = 2, x = - 15

4 x \geq - 20

18 x^{4} - 32 x^{2}

a^{2} - 121

x^{2} - 16 x + 48 < 0

\frac{5}{8}

What are the solutions to the equation 3x^2+13x-28=2x^2+2 ?
The solutions to the equation 3x^2+13x-28=2x^2+2 are x=2,x=-15
Find the zeros of 3x^2+13x-28=2x^2+2
The zeros of 3x^2+13x-28=2x^2+2 are x=2,x=-15