Solve 20x^2+10x-20=0

$$20 { x }^{ 2 } +10x-20=0$$

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=\frac{-\sqrt{17}-1}{4}\approx -1.280776406$$

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.

$$20x^{2}+10x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 10 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 20\left(-20\right)}}{2\times 20}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 20\left(-20\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-10±\sqrt{100-80\left(-20\right)}}{2\times 20}$$

Multiply -80 times -20.

$$x=\frac{-10±\sqrt{100+1600}}{2\times 20}$$

Add 100 to 1600.

$$x=\frac{-10±\sqrt{1700}}{2\times 20}$$

Take the square root of 1700.

$$x=\frac{-10±10\sqrt{17}}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-10±10\sqrt{17}}{40}$$

Now solve the equation x=\frac{-10±10\sqrt{17}}{40} when ± is plus. Add -10 to 10\sqrt{17}.

$$x=\frac{10\sqrt{17}-10}{40}$$

Divide -10+10\sqrt{17} by 40.

$$x=\frac{\sqrt{17}-1}{4}$$

Now solve the equation x=\frac{-10±10\sqrt{17}}{40} when ± is minus. Subtract 10\sqrt{17} from -10.

$$x=\frac{-10\sqrt{17}-10}{40}$$

Divide -10-10\sqrt{17} by 40.

$$x=\frac{-\sqrt{17}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$

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.

$$20x^{2}+10x-20=0$$

Add 20 to both sides of the equation.

$$20x^{2}+10x-20-\left(-20\right)=-\left(-20\right)$$

Subtracting -20 from itself leaves 0.

$$20x^{2}+10x=-\left(-20\right)$$

Subtract -20 from 0.

$$20x^{2}+10x=20$$

Divide both sides by 20.

$$\frac{20x^{2}+10x}{20}=\frac{20}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{10}{20}x=\frac{20}{20}$$

Reduce the fraction \frac{10}{20} to lowest terms by extracting and canceling out 10.

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

Divide 20 by 20.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=1+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}$$

Add 1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{17}}{4} x+\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$