The quadratic formula is x= -b +/- √b²-4ac / 2a

In this problem,

a=6
b=4
c=-3

Now, we can plug this into the formula:

x= -4 +/- √4²-(4)(6)(-3) / (2)(6)
x= -4 +/- √16+72 / 12
x= -4 +/- √88 /12
x= -4 +/- 2√22 /12
x= -2 +/- √22 / 6

So,

x= -2 + √22 / 6
x= -2 - √22 / 6

8 0

3 years ago

Read 2 more answers

hat are the solutions to the equation |x| = 2? Choose all answers that are correct. A. B. C. D.

zalisa [80]

Easy, so absolute value makes anything inside into positive
|x|=2
therefor x=-2 or x=2 since it would be made posotive

answer is -2 and 2

6 0

3 years ago

Read 2 more answers

How do I go about solving (27x^3/8y^9)^5/3? And what is the role of the numerator and denominator?

MrRissso [65]

$\left( \frac{27x^3}{8y^9}\right)^ \frac{5}{3} \\\\\\ =\left( \frac{(3x)^3}{(2y^3)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(3x)^{3 \times \frac{5}{3} }}{(2y^3)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(3x)^5}{(2y^3)^{5 }} \\\\\\ =\frac{243x^5}{32y^{15}}$

Now, If the exponent was negative like you asked....

$\left( \frac{27x^3}{8y^9}\right)^ {-\frac{5}{3}} \\\\\\ =\left( \frac{8y^9}{27x^3}\right)^ {\frac{5}{3}}\\\\\\ =\left( \frac{(2y^3)^3}{(3x)^3}\right)^ \frac{5}{3} \\\\\\ = \frac{(2y^3)^{3 \times \frac{5}{3} }}{(3x)^{3 \times \frac{5}{3} }} \\\\\\ =\frac{(2y^3)^{5 }}{(3x)^5} \\\\\\ =\frac{32y^{15}}{243x^5}$

5 0

3 years ago

Find the H.C.F of 48,60 and 96. please no links . please step by step explanation​

Find the H.C.F of 48,60 and 96. please no links . please step by step explanation