Solve 340x^2+80x-60=0

$$340 { x }^{ 2 } +80x-60 = 0$$

Solve for x

$$x=\frac{\sqrt{55}-2}{17}\approx 0.318599911$$

$$x=\frac{-\sqrt{55}-2}{17}\approx -0.553894029$$

Steps Using the Quadratic Formula

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.

$$340x^{2}+80x-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 340 for a, 80 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-80±\sqrt{80^{2}-4\times 340\left(-60\right)}}{2\times 340}$$

Square 80.

$$x=\frac{-80±\sqrt{6400-4\times 340\left(-60\right)}}{2\times 340}$$

Multiply -4 times 340.

$$x=\frac{-80±\sqrt{6400-1360\left(-60\right)}}{2\times 340}$$

Multiply -1360 times -60.

$$x=\frac{-80±\sqrt{6400+81600}}{2\times 340}$$

Add 6400 to 81600.

$$x=\frac{-80±\sqrt{88000}}{2\times 340}$$

Take the square root of 88000.

$$x=\frac{-80±40\sqrt{55}}{2\times 340}$$

Multiply 2 times 340.

$$x=\frac{-80±40\sqrt{55}}{680}$$

Now solve the equation x=\frac{-80±40\sqrt{55}}{680} when ± is plus. Add -80 to 40\sqrt{55}.

$$x=\frac{40\sqrt{55}-80}{680}$$

Divide -80+40\sqrt{55} by 680.

$$x=\frac{\sqrt{55}-2}{17}$$

Now solve the equation x=\frac{-80±40\sqrt{55}}{680} when ± is minus. Subtract 40\sqrt{55} from -80.

$$x=\frac{-40\sqrt{55}-80}{680}$$

Divide -80-40\sqrt{55} by 680.

$$x=\frac{-\sqrt{55}-2}{17}$$

The equation is now solved.

$$x=\frac{\sqrt{55}-2}{17} x=\frac{-\sqrt{55}-2}{17}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$340x^{2}+80x-60=0$$

Add 60 to both sides of the equation.

$$340x^{2}+80x-60-\left(-60\right)=-\left(-60\right)$$

Subtracting -60 from itself leaves 0.

$$340x^{2}+80x=-\left(-60\right)$$

Subtract -60 from 0.

$$340x^{2}+80x=60$$

Divide both sides by 340.

$$\frac{340x^{2}+80x}{340}=\frac{60}{340}$$

Dividing by 340 undoes the multiplication by 340.

$$x^{2}+\frac{80}{340}x=\frac{60}{340}$$

Reduce the fraction \frac{80}{340} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{4}{17}x=\frac{60}{340}$$

Reduce the fraction \frac{60}{340} to lowest terms by extracting and canceling out 20.

$$x^{2}+\frac{4}{17}x=\frac{3}{17}$$

Divide \frac{4}{17}, the coefficient of the x term, by 2 to get \frac{2}{17}. Then add the square of \frac{2}{17} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{17}x+\left(\frac{2}{17}\right)^{2}=\frac{3}{17}+\left(\frac{2}{17}\right)^{2}$$

Square \frac{2}{17} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{17}x+\frac{4}{289}=\frac{3}{17}+\frac{4}{289}$$

Add \frac{3}{17} to \frac{4}{289} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{17}x+\frac{4}{289}=\frac{55}{289}$$

Factor x^{2}+\frac{4}{17}x+\frac{4}{289}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{17}\right)^{2}=\frac{55}{289}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{17}\right)^{2}}=\sqrt{\frac{55}{289}}$$

Simplify.

$$x+\frac{2}{17}=\frac{\sqrt{55}}{17} x+\frac{2}{17}=-\frac{\sqrt{55}}{17}$$

Subtract \frac{2}{17} from both sides of the equation.

$$x=\frac{\sqrt{55}-2}{17} x=\frac{-\sqrt{55}-2}{17}$$