Solve (2k-3)^2-4(3-2k)<0

$$( 2 k - 3 ) ^ { 2 } - 4 ( 3 - 2 k ) < 0$$

Solve for k

$$k\in \left(-\frac{1}{2},\frac{3}{2}\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2k-3\right)^{2}.

$$4k^{2}-12k+9-4\left(3-2k\right)<0$$

Use the distributive property to multiply -4 by 3-2k.

$$4k^{2}-12k+9-12+8k<0$$

Subtract 12 from 9 to get -3.

$$4k^{2}-12k-3+8k<0$$

Combine -12k and 8k to get -4k.

$$4k^{2}-4k-3<0$$

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.

$$4k^{2}-4k-3=0$$

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, -4 for b, and -3 for c in the quadratic formula.

$$k=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Do the calculations.

$$k=\frac{4±8}{8}$$

Solve the equation k=\frac{4±8}{8} when ± is plus and when ± is minus.

$$k=\frac{3}{2} k=-\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(k-\frac{3}{2}\right)\left(k+\frac{1}{2}\right)<0$$

For the product to be negative, k-\frac{3}{2} and k+\frac{1}{2} have to be of the opposite signs. Consider the case when k-\frac{3}{2} is positive and k+\frac{1}{2} is negative.

$$k-\frac{3}{2}>0 k+\frac{1}{2}<0$$

This is false for any k.

$$k\in \emptyset$$

Consider the case when k+\frac{1}{2} is positive and k-\frac{3}{2} is negative.

$$k+\frac{1}{2}>0 k-\frac{3}{2}<0$$

The solution satisfying both inequalities is k\in \left(-\frac{1}{2},\frac{3}{2}\right).

$$k\in \left(-\frac{1}{2},\frac{3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$k\in \left(-\frac{1}{2},\frac{3}{2}\right)$$