Question

how can we solve x^+5x-300=0

Answer 1

x^2+5x-300=0 or, x^2+20x-15x-300=0 or, x(x+20)-15(x+20)=0 or, (x+20)(x-15)=0 or x= -20,15 ANSWER THANKS

Answer 2

For this eq...break the middle into such a part that the multiply of first and third equal to middle...
$x {}^{2} + 20x - 15 x - 300 = 0 \\ x(x + 20) - 15(x + 20) = 0\\ (x - 15)(x + 20) = 0 \\ x = 15 \: \: x = - 20$