Solve 4x^2-80x=188

$$4 { x }^{ 2 } -80x=188$$

Solve for x

$$x=7\sqrt{3}+10\approx 22.124355653$$

$$x=10-7\sqrt{3}\approx -2.124355653$$

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.

$$4x^{2}-80x=188$$

Subtract 188 from both sides of the equation.

$$4x^{2}-80x-188=188-188$$

Subtracting 188 from itself leaves 0.

$$4x^{2}-80x-188=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -80 for b, and -188 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 4\left(-188\right)}}{2\times 4}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 4\left(-188\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-80\right)±\sqrt{6400-16\left(-188\right)}}{2\times 4}$$

Multiply -16 times -188.

$$x=\frac{-\left(-80\right)±\sqrt{6400+3008}}{2\times 4}$$

Add 6400 to 3008.

$$x=\frac{-\left(-80\right)±\sqrt{9408}}{2\times 4}$$

Take the square root of 9408.

$$x=\frac{-\left(-80\right)±56\sqrt{3}}{2\times 4}$$

The opposite of -80 is 80.

$$x=\frac{80±56\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{80±56\sqrt{3}}{8}$$

Now solve the equation x=\frac{80±56\sqrt{3}}{8} when ± is plus. Add 80 to 56\sqrt{3}.

$$x=\frac{56\sqrt{3}+80}{8}$$

Divide 80+56\sqrt{3} by 8.

$$x=7\sqrt{3}+10$$

Now solve the equation x=\frac{80±56\sqrt{3}}{8} when ± is minus. Subtract 56\sqrt{3} from 80.

$$x=\frac{80-56\sqrt{3}}{8}$$

Divide 80-56\sqrt{3} by 8.

$$x=10-7\sqrt{3}$$

The equation is now solved.

$$x=7\sqrt{3}+10 x=10-7\sqrt{3}$$

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.

$$4x^{2}-80x=188$$

Divide both sides by 4.

$$\frac{4x^{2}-80x}{4}=\frac{188}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{80}{4}\right)x=\frac{188}{4}$$

Divide -80 by 4.

$$x^{2}-20x=\frac{188}{4}$$

Divide 188 by 4.

$$x^{2}-20x=47$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=47+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=47+100$$

Add 47 to 100.

$$x^{2}-20x+100=147$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=147$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{147}$$

Simplify.

$$x-10=7\sqrt{3} x-10=-7\sqrt{3}$$

Add 10 to both sides of the equation.

$$x=7\sqrt{3}+10 x=10-7\sqrt{3}$$