Solve x^2+4x^2+8x=357

$$x ^ { 2 } + 4 x ^ { 2 } + 8 x = 357$$

Solve for x

$$x = \frac{\sqrt{1801} - 4}{5} \approx 7.68763807$$

$$x=\frac{-\sqrt{1801}-4}{5}\approx -9.28763807$$

Steps Using the Quadratic Formula

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+8x=357$$

Subtract 357 from both sides.

$$5x^{2}+8x-357=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 8 for b, and -357 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 5\left(-357\right)}}{2\times 5}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 5\left(-357\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-8±\sqrt{64-20\left(-357\right)}}{2\times 5}$$

Multiply -20 times -357.

$$x=\frac{-8±\sqrt{64+7140}}{2\times 5}$$

Add 64 to 7140.

$$x=\frac{-8±\sqrt{7204}}{2\times 5}$$

Take the square root of 7204.

$$x=\frac{-8±2\sqrt{1801}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-8±2\sqrt{1801}}{10}$$

Now solve the equation x=\frac{-8±2\sqrt{1801}}{10} when ± is plus. Add -8 to 2\sqrt{1801}.

$$x=\frac{2\sqrt{1801}-8}{10}$$

Divide -8+2\sqrt{1801} by 10.

$$x=\frac{\sqrt{1801}-4}{5}$$

Now solve the equation x=\frac{-8±2\sqrt{1801}}{10} when ± is minus. Subtract 2\sqrt{1801} from -8.

$$x=\frac{-2\sqrt{1801}-8}{10}$$

Divide -8-2\sqrt{1801} by 10.

$$x=\frac{-\sqrt{1801}-4}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{1801}-4}{5} x=\frac{-\sqrt{1801}-4}{5}$$

Steps for Completing the Square

Combine x^{2} and 4x^{2} to get 5x^{2}.

$$5x^{2}+8x=357$$

Divide both sides by 5.

$$\frac{5x^{2}+8x}{5}=\frac{357}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{8}{5}x=\frac{357}{5}$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{5}x+\left(\frac{4}{5}\right)^{2}=\frac{357}{5}+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{357}{5}+\frac{16}{25}$$

Add \frac{357}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{5}x+\frac{16}{25}=\frac{1801}{25}$$

Factor x^{2}+\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{5}\right)^{2}=\frac{1801}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{5}\right)^{2}}=\sqrt{\frac{1801}{25}}$$

Simplify.

$$x+\frac{4}{5}=\frac{\sqrt{1801}}{5} x+\frac{4}{5}=-\frac{\sqrt{1801}}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

$$x=\frac{\sqrt{1801}-4}{5} x=\frac{-\sqrt{1801}-4}{5}$$