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

Question

Nathan Axel · Accepted Answer

See a solution process below:

One way to solve the quadratic is to factor it as: #(x - 2)(x + 5) = 0# Now, we can solve each term on the left for #0# to find the solutions: First Solution: #x - 2 = 0# #x - 2 + color(red)(2) = 0 + color(red)(2)# #x - 0 = 2# #x = 2# Option 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# The quadratic equation can also be used to solve this issue: According to the quadratic formula, 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))# Replacing: #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#

Ethan Adkins · Answer

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

$$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

In this equation, $a = 1$, $b = 3$, and $c = -10$.

Substitute these values into the quadratic formula:

$$x = \frac{{-3 \pm \sqrt{{(3)^2 - 4(1)(-10)}}}}{{2(1)}}$$

$$x = \frac{{-3 \pm \sqrt{{9 + 40}}}}{{2}}$$

$$x = \frac{{-3 \pm \sqrt{{49}}}}{{2}}$$

$$x = \frac{{-3 \pm 7}}{{2}}$$

This gives two possible solutions:

$$x_1 = \frac{{-3 + 7}}{{2}} = 2$$

$$x_2 = \frac{{-3 - 7}}{{2}} = -5$$

Therefore, the solutions to the equation $x^2 + 3x - 10 = 0$ are $x = 2$ and $x = -5$.