Solve by using the quadratic formula?

3x^2+4x+10=0

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2 Answers
Jun 18, 2018

See a solution process below:

Explanation:

The quadratic formula states:

For color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0, the values of x which are the solutions to the equation are given by:

x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))

Substituting:

color(red)(3) for color(red)(a)

color(blue)(4) for color(blue)(b)

color(green)(10) for color(green)(c) gives:

x = (-color(blue)(4) +- sqrt(color(blue)(4)^2 - (4 * color(red)(3) * color(green)(10))))/(2 * color(red)(3))

x = (-color(blue)(4) +- sqrt(16 - 120))/6

x = (-color(blue)(4) +- sqrt(-104))/6

x = (-color(blue)(4) +- sqrt(4 xx -26))/6

x = (-color(blue)(4) +- sqrt(4)sqrt(-26))/6

x = (-color(blue)(4) +- 2sqrt(-26))/6

Jun 18, 2018

No real solution.

Explanation:

The quadratic formular is x= (-b+- sqrt(b^2-4ac))/(2a) for the equation color(red)(a)x^2+color(blue)(b)x+color(orange)(c)=0

Therefore, in your case (color(red)(3)x^2+color(blue)(4)x+color(orange)(10)=0)
a=color(red)(3)
b=color(blue)(4)
c=color(orange)(10)

Using the formular, we get:
x= (-color(blue)(4)+- sqrt(color(blue)(4)^2-4*color(red)(3)*color(orange)(10)))/(2*color(red)(3))
x= (-4+- sqrt(16-120))/(6)
x=-2/3+-sqrt(color(green)(-104))/6

Since the radicand (color(green)(-104)) is negative, this equation has no real solutions for x.