Question

solve each of the following by quadratic equation (2x-3)(3x+1)=0​

Answer 1

(2x-3)(3x+1)=0

multiply (2x-3) seperately with (3x+1)

2x(3x+1)-3(3x+1)=0

6x^2+2x-9x-3=0

6x^2-7x-3=0

choose factors in such a way that if we add those we must get -7x and if we multiply we must get -18(coefficient of x^2×constant 6×-3=-18)

6x^2-9x+2x-3=0

3x(2x-3)+1(2x-3)=0

(2x-3)+(3x+1)=0

1.(2x-3)=0

2x=3

x=3/2

2.(3 x+1)=0

3x=-1

x=-1/3

hope it helps you ,☺

Answer 2

I hope it's help ful to u. ❤️❤️❤️❤️❤️❤️❤️❤️