One solution is to factor the quadratic as:

#(x - 2)(x + 5) = 0#

Now, we can solve each term on the left for #0# to find the solutions:

**Solution 1:**

#x - 2 = 0#

#x - 2 + color(red)(2) = 0 + color(red)(2)#

#x - 0 = 2#

#x = 2#

**Solution 2:**

#x + 5 = 0#

#x + 5 - color(red)(5) = 0 - color(red)(5)#

#x + 0 = -5#

#x = -5#

**he Solutions Are:** #x = 2# and #x = -5#

We can also 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)(1)# for #color(red)(a)#

#color(blue)(3)# for #color(blue)(b)#

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

#x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(1) * color(green)(-10))))/(2 * color(red)(1))#

#x = (-color(blue)(3) +- sqrt(9 - (-40)))/2#

#x = (-color(blue)(3) +- sqrt(9 + 40))/2#

#x = (-color(blue)(3) - sqrt(9 + 40))/2# and #x = (-color(blue)(3) + sqrt(9 + 40))/2#

#x = (-color(blue)(3) - sqrt(49))/2# and #x = (-color(blue)(3) + sqrt(49))/2#

#x = (-color(blue)(3) - 7)/2# and #x = (-color(blue)(3) + 7)/2#

#x = -10/2# and #x = 4/2#

#x = -5# and #x = 2#