How do you solve #3x^2+13x-10=0#?

3 Answers
Jul 22, 2017

Answer:

See a solution process below:

Explanation:

We can use the quadratic formula to solve this problem.

The quadratic formula states:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting:

#color(red)(3)# for #color(red)(a)#

#color(blue)(13)# for #color(blue)(b)#

#color(green)(-10)# for #color(green)(c)# gives:

#x = (-color(blue)(13) +- sqrt(color(blue)(13)^2 - (4 * color(red)(3) * color(green)(-10))))/(2 * color(red)(3))#

#x = (-color(blue)(13) +- sqrt(169 - (-120)))/6#

#x = (-color(blue)(13) +- sqrt(169 + 120))/6#

#x = (-color(blue)(13) +- sqrt(289))/6#

#x = (-color(blue)(13) + 17)/6# and #x = (-color(blue)(13) - 17)/6#

#x = 4/6# and #x = -30/6#

#x = 2/3# and #x = -5#

Jul 23, 2017

Answer:

#2/3# and - 5

Explanation:

#y = 3x^2 + 13x - 10 = 0#
Use the new Transforming Method (Google Search):
Transformed equation:
#y' = x^2 + 13x - 30 = 0#
Proceeding. Find 2 real roots of y', then. divide them by a = 3.
Find 2 real roots knowing the sum (-b = -13) and the product (ac = -30). They are: 2 and - 15.
Back to y, the 2 real roots are: #x1 = 2/a = 2/3# and
#x2 = -15/a = -15/3 = -5#

Jul 23, 2017

Answer:

#x=2/3" "# and #" "x=-5#

Explanation:

Given:

#3x^2+13x-10=0#

Use an AC method to factor the quadratic:

Look for a pair of factors of #AC=3*10 = 30# with difference #B=13#

The pair #15,2# works.

Use this pair to split the middle term and factor by grouping:

#0 = 3x^2+13x-10#

#color(white)(0) = (3x^2+15x)-(2x+10)#

#color(white)(0) = 3x(x+5)-2(x+5)#

#color(white)(0) = (3x-2)(x+5)#

Hence zeros:

#x=2/3" "# and #" "x=-5#