Solve a^2+2a+4a^2-5

$$a ^ { 2 } + 2 a + 4 a ^ { 2 } - 5$$

Factor

$$5\left(a-\frac{-\sqrt{26}-1}{5}\right)\left(a-\frac{\sqrt{26}-1}{5}\right)$$

Steps Using the Quadratic Formula

Combine a^{2} and 4a^{2} to get 5a^{2}.

$$factor(5a^{2}+2a-5)$$

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.

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

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.

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

Square 2.

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

Multiply -4 times 5.

$$a=\frac{-2±\sqrt{4-20\left(-5\right)}}{2\times 5}$$

Multiply -20 times -5.

$$a=\frac{-2±\sqrt{4+100}}{2\times 5}$$

Add 4 to 100.

$$a=\frac{-2±\sqrt{104}}{2\times 5}$$

Take the square root of 104.

$$a=\frac{-2±2\sqrt{26}}{2\times 5}$$

Multiply 2 times 5.

$$a=\frac{-2±2\sqrt{26}}{10}$$

Now solve the equation a=\frac{-2±2\sqrt{26}}{10} when ± is plus. Add -2 to 2\sqrt{26}.

$$a=\frac{2\sqrt{26}-2}{10}$$

Divide -2+2\sqrt{26} by 10.

$$a=\frac{\sqrt{26}-1}{5}$$

Now solve the equation a=\frac{-2±2\sqrt{26}}{10} when ± is minus. Subtract 2\sqrt{26} from -2.

$$a=\frac{-2\sqrt{26}-2}{10}$$

Divide -2-2\sqrt{26} by 10.

$$a=\frac{-\sqrt{26}-1}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{26}}{5} for x_{1} and \frac{-1-\sqrt{26}}{5} for x_{2}.

$$5a^{2}+2a-5=5\left(a-\frac{\sqrt{26}-1}{5}\right)\left(a-\frac{-\sqrt{26}-1}{5}\right)$$

Evaluate

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

Solution Steps

Combine a^{2} and 4a^{2} to get 5a^{2}.

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