Solve a^2+b3-4a^2/6sqrt{7a}=frac{sqrt{2}}{4}

$$\frac { a ^ { 2 } + b 3 - 4 a ^ { 2 } } { 6 \sqrt { 7 } a } = \frac { \sqrt { 2 } } { 4 }$$

Solve for b_3

$$b_{3}=\frac{3a\left(2a+\sqrt{14}\right)}{2}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$\frac{2}{21}a^{-1}\times 7^{\frac{1}{2}}\left(a^{2}+b_{3}-4a^{2}\right)=\sqrt{2}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$\frac{2}{21}a^{-1}\times 7^{\frac{1}{2}}\left(-3a^{2}+b_{3}\right)=\sqrt{2}$$

Use the distributive property to multiply \frac{2}{21}a^{-1}\times 7^{\frac{1}{2}} by -3a^{2}+b_{3}.

$$-\frac{2}{7}\times 7^{\frac{1}{2}}a+\frac{2}{21}a^{-1}\times 7^{\frac{1}{2}}b_{3}=\sqrt{2}$$

Add \frac{2}{7}\times 7^{\frac{1}{2}}a to both sides.

$$\frac{2}{21}a^{-1}\times 7^{\frac{1}{2}}b_{3}=\sqrt{2}+\frac{2}{7}\times 7^{\frac{1}{2}}a$$

Reorder the terms.

$$\frac{2}{21}\sqrt{7}\times \frac{1}{a}b_{3}=\frac{2}{7}\sqrt{7}a+\sqrt{2}$$

Multiply both sides of the equation by 21a, the least common multiple of 21,a,7.

$$\frac{2}{21}\sqrt{7}\times 21\times 1b_{3}=\frac{2}{7}\sqrt{7}a\times 21a+21a\sqrt{2}$$

Multiply \frac{2}{21} and 21 to get 2.

$$2\sqrt{7}\times 1b_{3}=\frac{2}{7}\sqrt{7}a\times 21a+21a\sqrt{2}$$

Multiply 2 and 1 to get 2.

$$2\sqrt{7}b_{3}=\frac{2}{7}\sqrt{7}a\times 21a+21a\sqrt{2}$$

Multiply a and a to get a^{2}.

$$2\sqrt{7}b_{3}=\frac{2}{7}\sqrt{7}a^{2}\times 21+21a\sqrt{2}$$

Multiply \frac{2}{7} and 21 to get 6.

$$2\sqrt{7}b_{3}=6\sqrt{7}a^{2}+21a\sqrt{2}$$

The equation is in standard form.

$$2\sqrt{7}b_{3}=6\sqrt{7}a^{2}+21\sqrt{2}a$$

Divide both sides by 2\sqrt{7}.

$$\frac{2\sqrt{7}b_{3}}{2\sqrt{7}}=\frac{6\sqrt{7}a^{2}+21\sqrt{2}a}{2\sqrt{7}}$$

Dividing by 2\sqrt{7} undoes the multiplication by 2\sqrt{7}.

$$b_{3}=\frac{6\sqrt{7}a^{2}+21\sqrt{2}a}{2\sqrt{7}}$$

Divide 6\sqrt{7}a^{2}+21a\sqrt{2} by 2\sqrt{7}.

$$b_{3}=\frac{3\sqrt{2}a\left(\sqrt{2}a+\sqrt{7}\right)}{2}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{\sqrt{48b_{3}+126}-3\sqrt{14}}{12}\text{, }&b_{3}\neq 0\text{ and }b_{3}\geq -\frac{21}{8}\\a=\frac{-\sqrt{48b_{3}+126}-3\sqrt{14}}{12}\text{, }&b_{3}\geq -\frac{21}{8}\end{matrix}\right.$$