Solve 2*a^2+11a+10=

$$2 \cdot a ^ { 2 } + 11 a + 10 =$$

Factor

$$2\left(a-\frac{-\sqrt{41}-11}{4}\right)\left(a-\frac{\sqrt{41}-11}{4}\right)$$

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.

$$2a^{2}+11a+10=0$$

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.

$$a=\frac{-11±\sqrt{11^{2}-4\times 2\times 10}}{2\times 2}$$

Square 11.

$$a=\frac{-11±\sqrt{121-4\times 2\times 10}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-11±\sqrt{121-8\times 10}}{2\times 2}$$

Multiply -8 times 10.

$$a=\frac{-11±\sqrt{121-80}}{2\times 2}$$

Add 121 to -80.

$$a=\frac{-11±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{-11±\sqrt{41}}{4}$$

Now solve the equation a=\frac{-11±\sqrt{41}}{4} when ± is plus. Add -11 to \sqrt{41}.

$$a=\frac{\sqrt{41}-11}{4}$$

Now solve the equation a=\frac{-11±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -11.

$$a=\frac{-\sqrt{41}-11}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-11+\sqrt{41}}{4} for x_{1} and \frac{-11-\sqrt{41}}{4} for x_{2}.

$$2a^{2}+11a+10=2\left(a-\frac{\sqrt{41}-11}{4}\right)\left(a-\frac{-\sqrt{41}-11}{4}\right)$$

Evaluate

$$2a^{2}+11a+10$$