Solve 3x^2 + x + 10 = 0. Round solutions to the nearest hundredth.

x ≈ -2.83 and x ≈ 0.83

No real solutions

x ≈ -2.01 and x ≈ 1.67

x ≈ -1.67 and x ≈ 2.01

Question

Solve 3x^2 + x + 10 = 0. Round solutions to the nearest hundredth.

 x ≈ -2.83 and x ≈ 0.83

 No real solutions

 x ≈ -2.01 and x ≈ 1.67

 x ≈ -1.67 and x ≈ 2.01

anonymous · Answer

@wio

anonymous · Answer

This is a quadratic polynomial so you will have to use the quadratic equation to solve for x

anonymous · Answer

Do you remember the quadratic equation?

anonymous · Answer

no im sorry

anonymous · Answer

$$x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$$

anonymous · Answer

$$ax^2+bx+c$$

anonymous · Answer

That is the form that your polynomial is in

anonymous · Answer

so how do i solve it?

anonymous · Answer

You use the values of a, b, and c in your quadratic equation and plug thenm into the quadratic formula

anonymous · Answer

ok, hold on a sec.

anonymous · Answer

x=-10+ then the symbol 10^2-4, thats all  can figure out..