Solve 5x^2-80/5x+20=0

$$\frac { 5 x ^ { 2 } - 80 } { 5 x + 20 } = 0$$

Solve for x

$$x=4$$

Steps Using Difference of Squares

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+4\right).

$$5x^{2}-80=0$$

Divide both sides by 5.

$$x^{2}-16=0$$

Consider x^{2}-16. Rewrite x^{2}-16 as x^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-4\right)\left(x+4\right)=0$$

To find equation solutions, solve x-4=0 and x+4=0.

$$x=4 x=-4$$

Variable x cannot be equal to -4.

$$x=4$$

Steps by Finding Square Root

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+4\right).

$$5x^{2}-80=0$$

Add 80 to both sides. Anything plus zero gives itself.

$$5x^{2}=80$$

Divide both sides by 5.

$$x^{2}=\frac{80}{5}$$

Divide 80 by 5 to get 16.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Variable x cannot be equal to -4.

$$x=4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by 5\left(x+4\right).

$$5x^{2}-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-80\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-80\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-80\right)}}{2\times 5}$$

Multiply -20 times -80.

$$x=\frac{0±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$x=\frac{0±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±40}{10}$$

Now solve the equation x=\frac{0±40}{10} when ± is plus. Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{0±40}{10} when ± is minus. Divide -40 by 10.

$$x=-4$$

The equation is now solved.

$$x=4 x=-4$$

Variable x cannot be equal to -4.

$$x=4$$