Solve 2x^2 + 19x - 145 = 0?

3 Answers
Mar 15, 2018

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(2)# for #color(red)(a)#

#color(blue)(19)# for #color(blue)(b)#

#color(green)(-145)# for #color(green)(c)# gives:

#x = (-color(blue)(19) +- sqrt(color(blue)(19)^2 - (4 * color(red)(2) * color(green)(-145))))/(2 * color(red)(2))#

#x = (-19 +- sqrt(361 - (8 * color(green)(-145))))/4#

#x = (-19 +- sqrt(361 - (-1160)))/4#

#x = (-19 +- sqrt(361 + 1160))/4#

#x = (-19 +- sqrt(1521))/4#

#x = (-19 - 39)/4# and #x = (-19 + 39)/4#

#x = (-58)/4# and #x = 20/4#

#x = -14.5# and #x = 5#

The Solution Set Is: #x = {-14.5, 5}#

Mar 15, 2018

See details below....

Explanation:

#2x^2+19x-145=0#

Start by factoring the the left side

#(2x + 29)(x-5)#

Then set factors equal to #0#

#2x + 29 = 0 or x-5=0#

#2x = 0 - 29 or x= 0 + 5#

#2x = -29 or x=5#

#x = (-29)/2 or x=5#

Mar 15, 2018

By using the quadratic formula, we find that x=5 and x=-14.5

Explanation:

The quadratic formula takes an equation that looks like this:

#ax^2+bx+c#

And plugs it into a formula that solves for x:

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

Based on our equation, we know the values of a, b, and c:

#a=2#
#b=19#
#c=-145#

#(-19+-sqrt(19^2-4(2xx-145)))/(2(2))#

#(-19+-sqrt(361+1160))/4 rArr (-19+-sqrt(1521))/4#

#(-19+-39)/4 rArr x=[5, -14.5]#