Solve 12x^2+34x+3

$$12 { x }^{ 2 } +34x+3$$

Factor

$$12\left(x-\frac{-\sqrt{253}-17}{12}\right)\left(x-\frac{\sqrt{253}-17}{12}\right)$$

Steps Using the Quadratic Formula

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.

$$12x^{2}+34x+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-34±\sqrt{34^{2}-4\times 12\times 3}}{2\times 12}$$

Square 34.

$$x=\frac{-34±\sqrt{1156-4\times 12\times 3}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-34±\sqrt{1156-48\times 3}}{2\times 12}$$

Multiply -48 times 3.

$$x=\frac{-34±\sqrt{1156-144}}{2\times 12}$$

Add 1156 to -144.

$$x=\frac{-34±\sqrt{1012}}{2\times 12}$$

Take the square root of 1012.

$$x=\frac{-34±2\sqrt{253}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-34±2\sqrt{253}}{24}$$

Now solve the equation x=\frac{-34±2\sqrt{253}}{24} when ± is plus. Add -34 to 2\sqrt{253}.

$$x=\frac{2\sqrt{253}-34}{24}$$

Divide -34+2\sqrt{253} by 24.

$$x=\frac{\sqrt{253}-17}{12}$$

Now solve the equation x=\frac{-34±2\sqrt{253}}{24} when ± is minus. Subtract 2\sqrt{253} from -34.

$$x=\frac{-2\sqrt{253}-34}{24}$$

Divide -34-2\sqrt{253} by 24.

$$x=\frac{-\sqrt{253}-17}{12}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-17+\sqrt{253}}{12} for x_{1} and \frac{-17-\sqrt{253}}{12} for x_{2}.

$$12x^{2}+34x+3=12\left(x-\frac{\sqrt{253}-17}{12}\right)\left(x-\frac{-\sqrt{253}-17}{12}\right)$$

Evaluate

$$12x^{2}+34x+3$$