Question

3xsquare-2x-5-10 = 0​ find roots is roots are real​

Answer 1

Answer:

3 -2-15=0

-2/3-5=0

(-2.591) (+1927)~0

~2.5941 ~-1.9274

~ 2.5941

Answer 2

Step-by-step explanation:

3x²-2x-15=0

Nature of the roots can be determined by discriminant

For the roots to be real D>=0

i.e, b²-4ac>=0

here b= -2

a= 3

c= -15

therefore. (-2)²-4(3)(-15)>=0

4+180>=0

184>=0

Since discriminant is greater than zero, two distinct roots exist.

the two roots can be found by

$\frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \: \: and \: \: \: \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

therefore they are

$\frac{ - ( - 2) + \sqrt{184} }{6} \: \: and \: \: \frac{ - ( - 2) - \sqrt{184} }{6}$

on simplifying we get,

$\frac{ 2 + \sqrt{184} }{6} \: \: and \: \: \frac{ 2 - \sqrt{184} }{6}$

$\frac{ 2 + 2 \sqrt{46} }{6} \: \: and \: \: \frac{ 2 - 2 \sqrt{46} }{6}$

$\frac{ 1 + 1 \sqrt{46} }{3} \: \: and \: \: \frac{ 1 - 1 \sqrt{46} }{3} \: are \: the \: final \: simplified \: roots$

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