Respuesta :

texaschic101
(5x + 6)(x - 1) - (2x^2 - 5x + 3) = 0
5x^2 - 5x + 6x - 6 - (2x^2 - 5x + 3) = 0
5x^2 + x - 6 - 2x^2 + 5x - 3 = 0
3x^2 + 6x - 9 = 0
3(x^2 + 2x - 3) = 0
3(x - 1)(x + 3) = 0

x - 1 = 0
x = 1

x + 3 = 0
x = -3

solution : x = 1 or x = -3
Firstly: multiply the two first parenthesis:
 
(5x+6)(x-1)=5x^2-5x+6x-6=5x^2+x-6



then: subtract the last parenthesis from what we have gotten:

5x^2+x-6 - (2x^2-5x+3) = 5x^2+x-6 - 2x^2+5x-3 = 3x^2+6x-9



then equate it to zero:
3x^2+6x-9=0


take 3 as a common factor:
3(x^2+2x-3)=0

x^2+2x-3=0



then solve this quadratic for x:

by factoring: (x+3)(x-1)=0

x=-3
x=1


Done!




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