Solve 95x^2+10x-5=0

$$95 { x }^{ 2 } +10x-5=0$$

Solve for x

$$x=\frac{2\sqrt{5}-1}{19}\approx 0.182743998$$

$$x=\frac{-2\sqrt{5}-1}{19}\approx -0.288007156$$

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.

$$95x^{2}+10x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 95 for a, 10 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 95\left(-5\right)}}{2\times 95}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 95\left(-5\right)}}{2\times 95}$$

Multiply -4 times 95.

$$x=\frac{-10±\sqrt{100-380\left(-5\right)}}{2\times 95}$$

Multiply -380 times -5.

$$x=\frac{-10±\sqrt{100+1900}}{2\times 95}$$

Add 100 to 1900.

$$x=\frac{-10±\sqrt{2000}}{2\times 95}$$

Take the square root of 2000.

$$x=\frac{-10±20\sqrt{5}}{2\times 95}$$

Multiply 2 times 95.

$$x=\frac{-10±20\sqrt{5}}{190}$$

Now solve the equation x=\frac{-10±20\sqrt{5}}{190} when ± is plus. Add -10 to 20\sqrt{5}.

$$x=\frac{20\sqrt{5}-10}{190}$$

Divide -10+20\sqrt{5} by 190.

$$x=\frac{2\sqrt{5}-1}{19}$$

Now solve the equation x=\frac{-10±20\sqrt{5}}{190} when ± is minus. Subtract 20\sqrt{5} from -10.

$$x=\frac{-20\sqrt{5}-10}{190}$$

Divide -10-20\sqrt{5} by 190.

$$x=\frac{-2\sqrt{5}-1}{19}$$

The equation is now solved.

$$x=\frac{2\sqrt{5}-1}{19} x=\frac{-2\sqrt{5}-1}{19}$$

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.

$$95x^{2}+10x-5=0$$

Add 5 to both sides of the equation.

$$95x^{2}+10x-5-\left(-5\right)=-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$95x^{2}+10x=-\left(-5\right)$$

Subtract -5 from 0.

$$95x^{2}+10x=5$$

Divide both sides by 95.

$$\frac{95x^{2}+10x}{95}=\frac{5}{95}$$

Dividing by 95 undoes the multiplication by 95.

$$x^{2}+\frac{10}{95}x=\frac{5}{95}$$

Reduce the fraction \frac{10}{95} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{2}{19}x=\frac{5}{95}$$

Reduce the fraction \frac{5}{95} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{2}{19}x=\frac{1}{19}$$

Divide \frac{2}{19}, the coefficient of the x term, by 2 to get \frac{1}{19}. Then add the square of \frac{1}{19} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{19}x+\left(\frac{1}{19}\right)^{2}=\frac{1}{19}+\left(\frac{1}{19}\right)^{2}$$

Square \frac{1}{19} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{19}x+\frac{1}{361}=\frac{1}{19}+\frac{1}{361}$$

Add \frac{1}{19} to \frac{1}{361} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{19}x+\frac{1}{361}=\frac{20}{361}$$

Factor x^{2}+\frac{2}{19}x+\frac{1}{361}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{19}\right)^{2}=\frac{20}{361}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{19}\right)^{2}}=\sqrt{\frac{20}{361}}$$

Simplify.

$$x+\frac{1}{19}=\frac{2\sqrt{5}}{19} x+\frac{1}{19}=-\frac{2\sqrt{5}}{19}$$

Subtract \frac{1}{19} from both sides of the equation.

$$x=\frac{2\sqrt{5}-1}{19} x=\frac{-2\sqrt{5}-1}{19}$$