Solve 3a^2+a-5

$$3 { a }^{ 2 } +a-5$$

Factor

$$3\left(a-\frac{-\sqrt{61}-1}{6}\right)\left(a-\frac{\sqrt{61}-1}{6}\right)$$

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.

$$3a^{2}+a-5=0$$

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.

$$a=\frac{-1±\sqrt{1^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-1±\sqrt{1-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$a=\frac{-1±\sqrt{1+60}}{2\times 3}$$

Add 1 to 60.

$$a=\frac{-1±\sqrt{61}}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{-1±\sqrt{61}}{6}$$

Now solve the equation a=\frac{-1±\sqrt{61}}{6} when ± is plus. Add -1 to \sqrt{61}.

$$a=\frac{\sqrt{61}-1}{6}$$

Now solve the equation a=\frac{-1±\sqrt{61}}{6} when ± is minus. Subtract \sqrt{61} from -1.

$$a=\frac{-\sqrt{61}-1}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{61}}{6} for x_{1} and \frac{-1-\sqrt{61}}{6} for x_{2}.

$$3a^{2}+a-5=3\left(a-\frac{\sqrt{61}-1}{6}\right)\left(a-\frac{-\sqrt{61}-1}{6}\right)$$

Evaluate

$$3a^{2}+a-5$$