Question

решите уравнение 3x^2+4x-16=(x-4)^2

Answer 1

$3 x^{2} +4x-16= x^{2} -8 x +16;$;
$2 x^{2} +12x-32=0;$
D=b^2-4ac=144-4*2*(-32)=400;
x1=$frac{-12+20}{4} =2$;
x2=$frac{-12-20}{4} =-8$