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Ber [7]
2 years ago
11

Solve 3x2 + x + 10 = 0. Round solutions to the nearest hundredth.

Mathematics
2 answers:
Kobotan [32]2 years ago
6 0
<span>3x2 + x + 10 = 0,  delta = 1² - 4.3.10= - 119 <0 so the answer is
b) </span><span>No real solutions</span>
ira [324]2 years ago
5 0

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.

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The slope intercept form is y=mx+b
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So; 5-(-7) / 0-4= -3
Then you use one of the points to find b; I’ll use 0 and 5 (the first number is x and the second number is y)
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