What are the solutions to the equation 2x^2+7x=-3 ?

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

Find the zeros of 2x^2+7x=-3

The zeros of 2x^2+7x=-3 are x=-1/2 ,x=-3

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2 x^{2} + 7 x = - 3

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

Solution steps

2 x^{2} + 7 x = - 3

Move

3

to the left side

2 x^{2} + 7 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 3 = 5

x_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 5}{2 \cdot 2}, x_{2} = \frac{- 7 - 5}{2 \cdot 2}

x = \frac{- 7 + 5}{2 \cdot 2} : - \frac{1}{2}

x = \frac{- 7 - 5}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

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

3 x^{2} - 12 x + 6 = 0

∣ x + 2 ∣ + ∣ x ∣ \leq 3

- 3 x - 5 < 10

6 x + 6 = 24

lo g_{e} (10)

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