Question

Solve 3x2 + 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

Answer 1

3x2 + x + 10 = 0,  delta = 1² - 4.3.10= - 119 <0 so the answer is
b) No real solutions

Answer 2

Answer:

No real solutions.

B is correct

Step-by-step explanation:

Given: $3x^2+x+10=0$

We are given a quadratic equation and need to solve for x.

Quadratic formula: $ax^2+bx+c=0$

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

For our equation, a=3, b=1 and c=10

Substitute the value of a, b and c into formula

$x=\dfrac{-1\pm\sqrt{1^2-4\cdot 3\cdot 10}}{2\cdot 3}$

$x=\dfrac{-1\pm\sqrt{-119}}{6}$

Here, Roots are imaginary because number inside the square root is negative.

Hence, No real solutions.