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Find the break even points of the expression $15x^2-26x-57=0$

Step-by-step Solution

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Final Answer

$x=3,\:x=-\frac{19}{15}$
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Step-by-step Solution

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Factor the trinomial $15x^2-26x-57$ of the form $ax^2+bx+c$, first, make the product of $15$ and $-57$

$\left(15\right)\left(-57\right)=-855$

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$\left(15\right)\left(-57\right)=-855$

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Learn how to solve classify algebraic expressions problems step by step online. Find the break even points of the expression 15x^2-26x+-57=0. Factor the trinomial 15x^2-26x-57 of the form ax^2+bx+c, first, make the product of 15 and -57. Now, find two numbers that multiplied give us -855 and add up to -26. Rewrite the original expression. Factor 15x^2-45x+19x-57 by the greatest common divisor 15.

Final Answer

$x=3,\:x=-\frac{19}{15}$

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Function Plot

Plotting: $15x^2-26x-57$

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Classify algebraic expressions

An algebraic expression can be classified as a monomial, binomial, trinomial or polynomial, depending on the number of terms.

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