Solve (b-2)^2-4*1*(b-3)geq0

$$( b - 2 ) ^ { 2 } - 4 \times 1 \times ( b - 3 ) \geq 0$$

Solve for b

$$b\in \mathrm{R}$$

Steps Using the Quadratic Formula

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

$$b^{2}-4b+4-4\times 1\left(b-3\right)\geq 0$$

Multiply 4 and 1 to get 4.

$$b^{2}-4b+4-4\left(b-3\right)\geq 0$$

Use the distributive property to multiply -4 by b-3.

$$b^{2}-4b+4-4b+12\geq 0$$

Combine -4b and -4b to get -8b.

$$b^{2}-8b+4+12\geq 0$$

Add 4 and 12 to get 16.

$$b^{2}-8b+16\geq 0$$

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.

$$b^{2}-8b+16=0$$

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

$$b=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$b=\frac{8±0}{2}$$

Solutions are the same.

$$b=4$$

Rewrite the inequality by using the obtained solutions.

$$\left(b-4\right)^{2}\geq 0$$

Inequality holds for b\in \mathrm{R}.

$$b\in \mathrm{R}$$