Solve (3a-2)^2-4(a-1)>0

$$( 3 a - 2 ) ^ { 2 } - 4 ( a - 1 ) > 0$$

Solve for a

$$a\in \mathrm{R}$$

Steps Using the Quadratic Formula

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

$$9a^{2}-12a+4-4\left(a-1\right)>0$$

Use the distributive property to multiply -4 by a-1.

$$9a^{2}-12a+4-4a+4>0$$

Combine -12a and -4a to get -16a.

$$9a^{2}-16a+4+4>0$$

Add 4 and 4 to get 8.

$$9a^{2}-16a+8>0$$

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.

$$9a^{2}-16a+8=0$$

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 9 for a, -16 for b, and 8 for c in the quadratic formula.

$$a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 9\times 8}}{2\times 9}$$

Do the calculations.

$$a=\frac{16±\sqrt{-32}}{18}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Expression 9a^{2}-16a+8 has the same sign for any a. To determine the sign, calculate the value of the expression for a=0.

$$9\times 0^{2}-16\times 0+8=8$$

The value of the expression 9a^{2}-16a+8 is always positive. Inequality holds for a\in \mathrm{R}.

$$a\in \mathrm{R}$$