Solve 5x^2+4x-20/2x=20

$$\frac{ 5 { x }^{ 2 } +4x-20 }{ 2x } =20$$

Solve for x

$$x = \frac{2 \sqrt{106} + 18}{5} \approx 7.718252056$$

$$x=\frac{18-2\sqrt{106}}{5}\approx -0.518252056$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$5x^{2}+4x-20=40x$$

Subtract 40x from both sides.

$$5x^{2}+4x-20-40x=0$$

Combine 4x and -40x to get -36x.

$$5x^{2}-36x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -36 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

Square -36.

$$x=\frac{-\left(-36\right)±\sqrt{1296-4\times 5\left(-20\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-36\right)±\sqrt{1296-20\left(-20\right)}}{2\times 5}$$

Multiply -20 times -20.

$$x=\frac{-\left(-36\right)±\sqrt{1296+400}}{2\times 5}$$

Add 1296 to 400.

$$x=\frac{-\left(-36\right)±\sqrt{1696}}{2\times 5}$$

Take the square root of 1696.

$$x=\frac{-\left(-36\right)±4\sqrt{106}}{2\times 5}$$

The opposite of -36 is 36.

$$x=\frac{36±4\sqrt{106}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{36±4\sqrt{106}}{10}$$

Now solve the equation x=\frac{36±4\sqrt{106}}{10} when ± is plus. Add 36 to 4\sqrt{106}.

$$x=\frac{4\sqrt{106}+36}{10}$$

Divide 36+4\sqrt{106} by 10.

$$x=\frac{2\sqrt{106}+18}{5}$$

Now solve the equation x=\frac{36±4\sqrt{106}}{10} when ± is minus. Subtract 4\sqrt{106} from 36.

$$x=\frac{36-4\sqrt{106}}{10}$$

Divide 36-4\sqrt{106} by 10.

$$x=\frac{18-2\sqrt{106}}{5}$$

The equation is now solved.

$$x=\frac{2\sqrt{106}+18}{5} x=\frac{18-2\sqrt{106}}{5}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$5x^{2}+4x-20=40x$$

Subtract 40x from both sides.

$$5x^{2}+4x-20-40x=0$$

Combine 4x and -40x to get -36x.

$$5x^{2}-36x-20=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$5x^{2}-36x=20$$

Divide both sides by 5.

$$\frac{5x^{2}-36x}{5}=\frac{20}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{36}{5}x=\frac{20}{5}$$

Divide 20 by 5.

$$x^{2}-\frac{36}{5}x=4$$

Divide -\frac{36}{5}, the coefficient of the x term, by 2 to get -\frac{18}{5}. Then add the square of -\frac{18}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{36}{5}x+\left(-\frac{18}{5}\right)^{2}=4+\left(-\frac{18}{5}\right)^{2}$$

Square -\frac{18}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{36}{5}x+\frac{324}{25}=4+\frac{324}{25}$$

Add 4 to \frac{324}{25}.

$$x^{2}-\frac{36}{5}x+\frac{324}{25}=\frac{424}{25}$$

Factor x^{2}-\frac{36}{5}x+\frac{324}{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}.

$$\left(x-\frac{18}{5}\right)^{2}=\frac{424}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{18}{5}\right)^{2}}=\sqrt{\frac{424}{25}}$$

Simplify.

$$x-\frac{18}{5}=\frac{2\sqrt{106}}{5} x-\frac{18}{5}=-\frac{2\sqrt{106}}{5}$$

Add \frac{18}{5} to both sides of the equation.

$$x=\frac{2\sqrt{106}+18}{5} x=\frac{18-2\sqrt{106}}{5}$$