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

Question

Genesis Allen · Accepted Answer

See a solution process below:

The quadratic formula can be applied to resolve this issue. According to the quadratic formula, 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)# Replacing: #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#

Hazel Anglin · Answer

#2/3# and - 5

#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#

Julia Adams · Answer

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

Given:

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

To factor the quadratic, apply the AC method:

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

The pair #15,2# works.

To divide the middle term and factor by grouping, use this pair:

#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)#

Thus, zeros:

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

Abigail Allen · Answer

undefined To solve the quadratic equation 3x^2 + 13x - 10 = 0, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where:
a = 3
b = 13
c = -10

Substitute these values into the formula:

x = (-(13) ± √((13)^2 - 4(3)(-10))) / (2(3))

Calculate inside the square root:

x = (-13 ± √(169 + 120)) / 6

x = (-13 ± √289) / 6

x = (-13 ± 17) / 6

This gives two possible solutions:

x₁ = (-13 + 17) / 6 = 4 / 6 = 2/3

x₂ = (-13 - 17) / 6 = -30 / 6 = -5

Therefore, the solutions are x = 2/3 and x = -5.