Calculate $\sqrt{5x-9}$ to the power of $2$ and get $5x-9$.
$$4\left(5x-9\right)=\left(14+51x\right)^{2}$$
Use the distributive property to multiply $4$ by $5x-9$.
$$20x-36=\left(14+51x\right)^{2}$$
Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(14+51x\right)^{2}$.
$$20x-36=196+1428x+2601x^{2}$$
Subtract $1428x$ from both sides.
$$20x-36-1428x=196+2601x^{2}$$
Combine $20x$ and $-1428x$ to get $-1408x$.
$$-1408x-36=196+2601x^{2}$$
Subtract $2601x^{2}$ from both sides.
$$-1408x-36-2601x^{2}=196$$
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.
$$-2601x^{2}-1408x-36=196$$
Subtract $196$ from both sides of the equation.
$$-2601x^{2}-1408x-36-196=196-196$$
Subtracting $196$ from itself leaves $0$.
$$-2601x^{2}-1408x-36-196=0$$
Subtract $196$ from $-36$.
$$-2601x^{2}-1408x-232=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2601$ for $a$, $-1408$ for $b$, and $-232$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
