Question

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

Answer 1

Answer:

C. X= -2.01 and x= 1.67

Step-by-step explanation:

$3x {}^{2} + x + 10 = 0 \\ 3x {}^{2} + 6x - 5x + 10 = 0 \\ 3x(x + 2) - 5(x + 2) \\ (x + 2)(3x - 5 )\\ x + 2 = 0 \: \: or \: \: 3x - 5 = 0 \\ x = - 2 \: \: or \: \: x = \frac{5}{3}$

Answer 2

ANSWER

B. No real solutions

EXPLANATION

The given equation is

$3 {x}^{2} + x + 10 = 0$

By comparing to

$a {x}^{2} + bx + c= 0$

We have a=3,b=1 and c=10.

We substitute these values into the formula

$D = {b}^{2} - 4ac$

to determine the nature of the roots.

$D = {1}^{2} - 4(3)(10)$

$D = 1 - 120$

$D = - 119$

The discriminant is negative.

This means that the given quadratic equation has no real roots.