Solve a^2+(4a+10)^2=64/25

$$a ^ { 2 } + ( 4 a + 10 ) ^ { 2 } = \frac { 64 } { 25 }$$

Solve for a

$$a=\frac{2\sqrt{353}i}{85}-\frac{40}{17}\approx -2.352941176+0.442077511i$$

$$a=-\frac{2\sqrt{353}i}{85}-\frac{40}{17}\approx -2.352941176-0.442077511i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4a+10\right)^{2}.

$$a^{2}+16a^{2}+80a+100=\frac{64}{25}$$

Combine a^{2} and 16a^{2} to get 17a^{2}.

$$17a^{2}+80a+100=\frac{64}{25}$$

Subtract \frac{64}{25} from both sides.

$$17a^{2}+80a+100-\frac{64}{25}=0$$

Subtract \frac{64}{25} from 100 to get \frac{2436}{25}.

$$17a^{2}+80a+\frac{2436}{25}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 17 for a, 80 for b, and \frac{2436}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-80±\sqrt{80^{2}-4\times 17\times \frac{2436}{25}}}{2\times 17}$$

Square 80.

$$a=\frac{-80±\sqrt{6400-4\times 17\times \frac{2436}{25}}}{2\times 17}$$

Multiply -4 times 17.

$$a=\frac{-80±\sqrt{6400-68\times \frac{2436}{25}}}{2\times 17}$$

Multiply -68 times \frac{2436}{25}.

$$a=\frac{-80±\sqrt{6400-\frac{165648}{25}}}{2\times 17}$$

Add 6400 to -\frac{165648}{25}.

$$a=\frac{-80±\sqrt{-\frac{5648}{25}}}{2\times 17}$$

Take the square root of -\frac{5648}{25}.

$$a=\frac{-80±\frac{4\sqrt{353}i}{5}}{2\times 17}$$

Multiply 2 times 17.

$$a=\frac{-80±\frac{4\sqrt{353}i}{5}}{34}$$

Now solve the equation a=\frac{-80±\frac{4\sqrt{353}i}{5}}{34} when ± is plus. Add -80 to \frac{4i\sqrt{353}}{5}.

$$a=\frac{\frac{4\sqrt{353}i}{5}-80}{34}$$

Divide -80+\frac{4i\sqrt{353}}{5} by 34.

$$a=\frac{2\sqrt{353}i}{85}-\frac{40}{17}$$

Now solve the equation a=\frac{-80±\frac{4\sqrt{353}i}{5}}{34} when ± is minus. Subtract \frac{4i\sqrt{353}}{5} from -80.

$$a=\frac{-\frac{4\sqrt{353}i}{5}-80}{34}$$

Divide -80-\frac{4i\sqrt{353}}{5} by 34.

$$a=-\frac{2\sqrt{353}i}{85}-\frac{40}{17}$$

The equation is now solved.

$$a=\frac{2\sqrt{353}i}{85}-\frac{40}{17} a=-\frac{2\sqrt{353}i}{85}-\frac{40}{17}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4a+10\right)^{2}.

$$a^{2}+16a^{2}+80a+100=\frac{64}{25}$$

Combine a^{2} and 16a^{2} to get 17a^{2}.

$$17a^{2}+80a+100=\frac{64}{25}$$

Subtract 100 from both sides.

$$17a^{2}+80a=\frac{64}{25}-100$$

Subtract 100 from \frac{64}{25} to get -\frac{2436}{25}.

$$17a^{2}+80a=-\frac{2436}{25}$$

Divide both sides by 17.

$$\frac{17a^{2}+80a}{17}=-\frac{\frac{2436}{25}}{17}$$

Dividing by 17 undoes the multiplication by 17.

$$a^{2}+\frac{80}{17}a=-\frac{\frac{2436}{25}}{17}$$

Divide -\frac{2436}{25} by 17.

$$a^{2}+\frac{80}{17}a=-\frac{2436}{425}$$

Divide \frac{80}{17}, the coefficient of the x term, by 2 to get \frac{40}{17}. Then add the square of \frac{40}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{80}{17}a+\left(\frac{40}{17}\right)^{2}=-\frac{2436}{425}+\left(\frac{40}{17}\right)^{2}$$

Square \frac{40}{17} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{80}{17}a+\frac{1600}{289}=-\frac{2436}{425}+\frac{1600}{289}$$

Add -\frac{2436}{425} to \frac{1600}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{80}{17}a+\frac{1600}{289}=-\frac{1412}{7225}$$

Factor a^{2}+\frac{80}{17}a+\frac{1600}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{40}{17}\right)^{2}=-\frac{1412}{7225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{40}{17}\right)^{2}}=\sqrt{-\frac{1412}{7225}}$$

Simplify.

$$a+\frac{40}{17}=\frac{2\sqrt{353}i}{85} a+\frac{40}{17}=-\frac{2\sqrt{353}i}{85}$$

Subtract \frac{40}{17} from both sides of the equation.

$$a=\frac{2\sqrt{353}i}{85}-\frac{40}{17} a=-\frac{2\sqrt{353}i}{85}-\frac{40}{17}$$