Решите неравенство 4x^2+10х-20<=(x+2)^2
4х^2+10х-20<=(X+2)^2</p>
4X^2+10x-20<=x^2+4x+4</p>
4x^2-x^2+10x-4x-20-4<=0</p>
3x^2+6x-24<=0</p>
D=b^2-4ac
D=36-4*3*(-24)=324
x1,2=(-b^2+-корень изD)/2a
x1=(-6+18)/(2*3)=2
x2=(-6-18)/(2*3)=-4
4x^2+10х-20<=(x+2)^2</span>
4x^2+10х-20<=x^2+4x+4</span>
3x^2+6x-24<=0</span>
x^2+2x-8<=0</span>
x1=-4 x2=2
xe[-4,2]