Solve 1+4(1-a)(a-2)<0

$$1 + 4 ( 1 - a ) ( a - 2 ) < 0$$

Solve for a

$$a\in \left(-\infty,\frac{3-\sqrt{2}}{2}\right)\cup \left(\frac{\sqrt{2}+3}{2},\infty\right)$$

Steps Using the Quadratic Formula

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

$$1+\left(4-4a\right)\left(a-2\right)<0$$

Use the distributive property to multiply 4-4a by a-2 and combine like terms.

$$1+12a-8-4a^{2}<0$$

Subtract 8 from 1 to get -7.

$$-7+12a-4a^{2}<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -7+12a-4a^{2} positive. Since -1 is negative, the inequality direction is changed.

$$7-12a+4a^{2}>0$$

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.

$$7-12a+4a^{2}=0$$

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, -12 for b, and 7 for c in the quadratic formula.

$$a=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 7}}{2\times 4}$$

Do the calculations.

$$a=\frac{12±4\sqrt{2}}{8}$$

Solve the equation a=\frac{12±4\sqrt{2}}{8} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{2}+3}{2} a=\frac{3-\sqrt{2}}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(a-\frac{\sqrt{2}+3}{2}\right)\left(a-\frac{3-\sqrt{2}}{2}\right)>0$$

For the product to be positive, a-\frac{\sqrt{2}+3}{2} and a-\frac{3-\sqrt{2}}{2} have to be both negative or both positive. Consider the case when a-\frac{\sqrt{2}+3}{2} and a-\frac{3-\sqrt{2}}{2} are both negative.

$$a-\frac{\sqrt{2}+3}{2}<0 a-\frac{3-\sqrt{2}}{2}<0$$

The solution satisfying both inequalities is a<\frac{3-\sqrt{2}}{2}.

$$a<\frac{3-\sqrt{2}}{2}$$

Consider the case when a-\frac{\sqrt{2}+3}{2} and a-\frac{3-\sqrt{2}}{2} are both positive.

$$a-\frac{3-\sqrt{2}}{2}>0 a-\frac{\sqrt{2}+3}{2}>0$$

The solution satisfying both inequalities is a>\frac{\sqrt{2}+3}{2}.

$$a>\frac{\sqrt{2}+3}{2}$$

The final solution is the union of the obtained solutions.

$$a<\frac{3-\sqrt{2}}{2}\text{; }a>\frac{\sqrt{2}+3}{2}$$