1)y=2x^2 + 5x
2)7=x^2 + 4x + 12
3)y=3x^2 + 4x - 6
1)=2) 2x^2 + 5x =x^2 + 4x - 12
x^2+x-12=0=(x+4)(x-3) so x=-4,3
1)=3)2x^2 + 5x =3x^2 + 4x - 6
so x^2-x-6= 0 so (x+2)(x-3)=0 so x=-2,3
2)=3)x^2 + 4x +12=3x^2+4x-6 so2x^2-18=0 so x^2=9 so x=3, -3
ans is x=3
y=x^2+4x+12=9+12+12=33