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
