Solve 4(4lambda^2-12lambda+9)-4(5lambda^2-10lambda-6lambda+12)>0

$$4 ( 4 \lambda ^ { 2 } - 12 \lambda + 9 ) - 4 ( 5 \lambda ^ { 2 } - 10 \lambda - 6 \lambda + 12 ) > 0$$

Solve for λ

$$\lambda \in \left(1,3\right)$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4 by 4\lambda ^{2}-12\lambda +9.

$$16\lambda ^{2}-48\lambda +36-4\left(5\lambda ^{2}-10\lambda -6\lambda +12\right)>0$$

Combine -10\lambda and -6\lambda to get -16\lambda .

$$16\lambda ^{2}-48\lambda +36-4\left(5\lambda ^{2}-16\lambda +12\right)>0$$

Use the distributive property to multiply -4 by 5\lambda ^{2}-16\lambda +12.

$$16\lambda ^{2}-48\lambda +36-20\lambda ^{2}+64\lambda -48>0$$

Combine 16\lambda ^{2} and -20\lambda ^{2} to get -4\lambda ^{2}.

$$-4\lambda ^{2}-48\lambda +36+64\lambda -48>0$$

Combine -48\lambda and 64\lambda to get 16\lambda .

$$-4\lambda ^{2}+16\lambda +36-48>0$$

Subtract 48 from 36 to get -12.

$$-4\lambda ^{2}+16\lambda -12>0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -4\lambda ^{2}+16\lambda -12 positive. Since -1 is negative, the inequality direction is changed.

$$4\lambda ^{2}-16\lambda +12<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4\lambda ^{2}-16\lambda +12=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -16 for b, and 12 for c in the quadratic formula.

$$\lambda =\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 12}}{2\times 4}$$

Do the calculations.

$$\lambda =\frac{16±8}{8}$$

Solve the equation \lambda =\frac{16±8}{8} when ± is plus and when ± is minus.

$$\lambda =3 \lambda =1$$

Rewrite the inequality by using the obtained solutions.

$$4\left(\lambda -3\right)\left(\lambda -1\right)<0$$

For the product to be negative, \lambda -3 and \lambda -1 have to be of the opposite signs. Consider the case when \lambda -3 is positive and \lambda -1 is negative.

$$\lambda -3>0 \lambda -1<0$$

This is false for any \lambda .

$$\lambda \in \emptyset$$

Consider the case when \lambda -1 is positive and \lambda -3 is negative.

$$\lambda -1>0 \lambda -3<0$$

The solution satisfying both inequalities is \lambda \in \left(1,3\right).

$$\lambda \in \left(1,3\right)$$

The final solution is the union of the obtained solutions.

$$\lambda \in \left(1,3\right)$$