Respuesta :

harpazo
2x^2+20x = -38

2x^2 + 20x + 38 = 0

Divide both sides by 2.

x^2 + 10x + 19 = 0

The quadratic formula works very well here.

Let a = 1, b = 10 and c = 19

x = [-b +- sqrt{b^2 - 4ac}]/(2a)

Plug those values into the formula to find two answers for x.

One is negative and the other positive.

Good luck.
miagwh
Move 38 to the left side of the equation by adding it to both sides.
2x ^2 + 20x + 38 = 0
Use the quadratic formula to find the solutions.
−b±√b^2 − 4 (ac)/2a
Substitute a = 2, b = 20, and c = 38 into the quadratic formula and solve for x.
−20±√20 ^2 − 4⋅(2⋅38)/2⋅2
Simplify. x = −5±√6
The result can be shown in both exact and approximate form.
x = −5±√6
x ≈ −2.55051025, −7.44948974

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