Solve 3x^2+80*2x-356=0

$$3 x ^ { 2 } + 80 \cdot 2 x - 356 = 0$$

Solve for x

$$x = \frac{2 \sqrt{1867} - 80}{3} \approx 2.139196934$$

$$x=\frac{-2\sqrt{1867}-80}{3}\approx -55.472530267$$

Steps Using the Quadratic Formula

Multiply 80 and 2 to get 160.

$$3x^{2}+160x-356=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 160 for b, and -356 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-160±\sqrt{160^{2}-4\times 3\left(-356\right)}}{2\times 3}$$

Square 160.

$$x=\frac{-160±\sqrt{25600-4\times 3\left(-356\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-160±\sqrt{25600-12\left(-356\right)}}{2\times 3}$$

Multiply -12 times -356.

$$x=\frac{-160±\sqrt{25600+4272}}{2\times 3}$$

Add 25600 to 4272.

$$x=\frac{-160±\sqrt{29872}}{2\times 3}$$

Take the square root of 29872.

$$x=\frac{-160±4\sqrt{1867}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-160±4\sqrt{1867}}{6}$$

Now solve the equation x=\frac{-160±4\sqrt{1867}}{6} when ± is plus. Add -160 to 4\sqrt{1867}.

$$x=\frac{4\sqrt{1867}-160}{6}$$

Divide -160+4\sqrt{1867} by 6.

$$x=\frac{2\sqrt{1867}-80}{3}$$

Now solve the equation x=\frac{-160±4\sqrt{1867}}{6} when ± is minus. Subtract 4\sqrt{1867} from -160.

$$x=\frac{-4\sqrt{1867}-160}{6}$$

Divide -160-4\sqrt{1867} by 6.

$$x=\frac{-2\sqrt{1867}-80}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{1867}-80}{3} x=\frac{-2\sqrt{1867}-80}{3}$$

Steps for Completing the Square

Multiply 80 and 2 to get 160.

$$3x^{2}+160x-356=0$$

Add 356 to both sides. Anything plus zero gives itself.

$$3x^{2}+160x=356$$

Divide both sides by 3.

$$\frac{3x^{2}+160x}{3}=\frac{356}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{160}{3}x=\frac{356}{3}$$

Divide \frac{160}{3}, the coefficient of the x term, by 2 to get \frac{80}{3}. Then add the square of \frac{80}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{160}{3}x+\left(\frac{80}{3}\right)^{2}=\frac{356}{3}+\left(\frac{80}{3}\right)^{2}$$

Square \frac{80}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{160}{3}x+\frac{6400}{9}=\frac{356}{3}+\frac{6400}{9}$$

Add \frac{356}{3} to \frac{6400}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{160}{3}x+\frac{6400}{9}=\frac{7468}{9}$$

Factor x^{2}+\frac{160}{3}x+\frac{6400}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{80}{3}\right)^{2}=\frac{7468}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{80}{3}\right)^{2}}=\sqrt{\frac{7468}{9}}$$

Simplify.

$$x+\frac{80}{3}=\frac{2\sqrt{1867}}{3} x+\frac{80}{3}=-\frac{2\sqrt{1867}}{3}$$

Subtract \frac{80}{3} from both sides of the equation.

$$x=\frac{2\sqrt{1867}-80}{3} x=\frac{-2\sqrt{1867}-80}{3}$$