Solve 2sqrt{n^2+n-12}=50-2n

$$2 \sqrt { n ^ { 2 } + n - 12 } = 50 - 2 n$$

Solve for n

$$n = \frac{637}{51} = 12\frac{25}{51} \approx 12.490196078$$

Solution Steps

Square both sides of the equation.

$$\left(2\sqrt{n^{2}+n-12}\right)^{2}=\left(50-2n\right)^{2}$$

Expand \left(2\sqrt{n^{2}+n-12}\right)^{2}.

$$2^{2}\left(\sqrt{n^{2}+n-12}\right)^{2}=\left(50-2n\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{n^{2}+n-12}\right)^{2}=\left(50-2n\right)^{2}$$

Calculate \sqrt{n^{2}+n-12} to the power of 2 and get n^{2}+n-12.

$$4\left(n^{2}+n-12\right)=\left(50-2n\right)^{2}$$

Use the distributive property to multiply 4 by n^{2}+n-12.

$$4n^{2}+4n-48=\left(50-2n\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(50-2n\right)^{2}.

$$4n^{2}+4n-48=2500-200n+4n^{2}$$

Add 200n to both sides.

$$4n^{2}+4n-48+200n=2500+4n^{2}$$

Combine 4n and 200n to get 204n.

$$4n^{2}+204n-48=2500+4n^{2}$$

Subtract 4n^{2} from both sides.

$$4n^{2}+204n-48-4n^{2}=2500$$

Combine 4n^{2} and -4n^{2} to get 0.

$$204n-48=2500$$

Add 48 to both sides.

$$204n=2500+48$$

Add 2500 and 48 to get 2548.

$$204n=2548$$

Divide both sides by 204.

$$n=\frac{2548}{204}$$

Reduce the fraction \frac{2548}{204} to lowest terms by extracting and canceling out 4.

$$n=\frac{637}{51}$$

Substitute \frac{637}{51} for n in the equation 2\sqrt{n^{2}+n-12}=50-2n.

$$2\sqrt{\left(\frac{637}{51}\right)^{2}+\frac{637}{51}-12}=50-2\times \frac{637}{51}$$

Simplify. The value n=\frac{637}{51} satisfies the equation.

$$\frac{1276}{51}=\frac{1276}{51}$$

Equation 2\sqrt{n^{2}+n-12}=50-2n has a unique solution.

$$n=\frac{637}{51}$$