Solve (6+16a)^2-4*4(2a^2-5)=0

$$( 6 + 16 a ) ^ { 2 } - 4 \times 4 ( 2 a ^ { 2 } - 5 ) = 0$$

Solve for a

$$a=\frac{\sqrt{262}i}{28}-\frac{3}{7}\approx -0.428571429+0.578086216i$$

$$a=-\frac{\sqrt{262}i}{28}-\frac{3}{7}\approx -0.428571429-0.578086216i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+16a\right)^{2}.

$$36+192a+256a^{2}-4\times 4\left(2a^{2}-5\right)=0$$

Multiply 4 and 4 to get 16.

$$36+192a+256a^{2}-16\left(2a^{2}-5\right)=0$$

Use the distributive property to multiply -16 by 2a^{2}-5.

$$36+192a+256a^{2}-32a^{2}+80=0$$

Combine 256a^{2} and -32a^{2} to get 224a^{2}.

$$36+192a+224a^{2}+80=0$$

Add 36 and 80 to get 116.

$$116+192a+224a^{2}=0$$

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.

$$224a^{2}+192a+116=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 224 for a, 192 for b, and 116 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-192±\sqrt{192^{2}-4\times 224\times 116}}{2\times 224}$$

Square 192.

$$a=\frac{-192±\sqrt{36864-4\times 224\times 116}}{2\times 224}$$

Multiply -4 times 224.

$$a=\frac{-192±\sqrt{36864-896\times 116}}{2\times 224}$$

Multiply -896 times 116.

$$a=\frac{-192±\sqrt{36864-103936}}{2\times 224}$$

Add 36864 to -103936.

$$a=\frac{-192±\sqrt{-67072}}{2\times 224}$$

Take the square root of -67072.

$$a=\frac{-192±16\sqrt{262}i}{2\times 224}$$

Multiply 2 times 224.

$$a=\frac{-192±16\sqrt{262}i}{448}$$

Now solve the equation a=\frac{-192±16\sqrt{262}i}{448} when ± is plus. Add -192 to 16i\sqrt{262}.

$$a=\frac{-192+16\sqrt{262}i}{448}$$

Divide -192+16i\sqrt{262} by 448.

$$a=\frac{\sqrt{262}i}{28}-\frac{3}{7}$$

Now solve the equation a=\frac{-192±16\sqrt{262}i}{448} when ± is minus. Subtract 16i\sqrt{262} from -192.

$$a=\frac{-16\sqrt{262}i-192}{448}$$

Divide -192-16i\sqrt{262} by 448.

$$a=-\frac{\sqrt{262}i}{28}-\frac{3}{7}$$

The equation is now solved.

$$a=\frac{\sqrt{262}i}{28}-\frac{3}{7} a=-\frac{\sqrt{262}i}{28}-\frac{3}{7}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+16a\right)^{2}.

$$36+192a+256a^{2}-4\times 4\left(2a^{2}-5\right)=0$$

Multiply 4 and 4 to get 16.

$$36+192a+256a^{2}-16\left(2a^{2}-5\right)=0$$

Use the distributive property to multiply -16 by 2a^{2}-5.

$$36+192a+256a^{2}-32a^{2}+80=0$$

Combine 256a^{2} and -32a^{2} to get 224a^{2}.

$$36+192a+224a^{2}+80=0$$

Add 36 and 80 to get 116.

$$116+192a+224a^{2}=0$$

Subtract 116 from both sides. Anything subtracted from zero gives its negation.

$$192a+224a^{2}=-116$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$224a^{2}+192a=-116$$

Divide both sides by 224.

$$\frac{224a^{2}+192a}{224}=-\frac{116}{224}$$

Dividing by 224 undoes the multiplication by 224.

$$a^{2}+\frac{192}{224}a=-\frac{116}{224}$$

Reduce the fraction \frac{192}{224} to lowest terms by extracting and canceling out 32.

$$a^{2}+\frac{6}{7}a=-\frac{116}{224}$$

Reduce the fraction \frac{-116}{224} to lowest terms by extracting and canceling out 4.

$$a^{2}+\frac{6}{7}a=-\frac{29}{56}$$

Divide \frac{6}{7}, the coefficient of the x term, by 2 to get \frac{3}{7}. Then add the square of \frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{6}{7}a+\left(\frac{3}{7}\right)^{2}=-\frac{29}{56}+\left(\frac{3}{7}\right)^{2}$$

Square \frac{3}{7} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{6}{7}a+\frac{9}{49}=-\frac{29}{56}+\frac{9}{49}$$

Add -\frac{29}{56} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{6}{7}a+\frac{9}{49}=-\frac{131}{392}$$

Factor a^{2}+\frac{6}{7}a+\frac{9}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{3}{7}\right)^{2}=-\frac{131}{392}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{3}{7}\right)^{2}}=\sqrt{-\frac{131}{392}}$$

Simplify.

$$a+\frac{3}{7}=\frac{\sqrt{262}i}{28} a+\frac{3}{7}=-\frac{\sqrt{262}i}{28}$$

Subtract \frac{3}{7} from both sides of the equation.

$$a=\frac{\sqrt{262}i}{28}-\frac{3}{7} a=-\frac{\sqrt{262}i}{28}-\frac{3}{7}$$