Solve 11x^2-80x+104=0

$$11 { x }^{ 2 } -80x+104=0$$

Solve for x

$$x = \frac{2 \sqrt{114} + 40}{11} \approx 5.577650591$$

$$x = \frac{40 - 2 \sqrt{114}}{11} \approx 1.695076681$$

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.

$$11x^{2}-80x+104=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, -80 for b, and 104 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 11\times 104}}{2\times 11}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 11\times 104}}{2\times 11}$$

Multiply -4 times 11.

$$x=\frac{-\left(-80\right)±\sqrt{6400-44\times 104}}{2\times 11}$$

Multiply -44 times 104.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4576}}{2\times 11}$$

Add 6400 to -4576.

$$x=\frac{-\left(-80\right)±\sqrt{1824}}{2\times 11}$$

Take the square root of 1824.

$$x=\frac{-\left(-80\right)±4\sqrt{114}}{2\times 11}$$

The opposite of -80 is 80.

$$x=\frac{80±4\sqrt{114}}{2\times 11}$$

Multiply 2 times 11.

$$x=\frac{80±4\sqrt{114}}{22}$$

Now solve the equation x=\frac{80±4\sqrt{114}}{22} when ± is plus. Add 80 to 4\sqrt{114}.

$$x=\frac{4\sqrt{114}+80}{22}$$

Divide 80+4\sqrt{114} by 22.

$$x=\frac{2\sqrt{114}+40}{11}$$

Now solve the equation x=\frac{80±4\sqrt{114}}{22} when ± is minus. Subtract 4\sqrt{114} from 80.

$$x=\frac{80-4\sqrt{114}}{22}$$

Divide 80-4\sqrt{114} by 22.

$$x=\frac{40-2\sqrt{114}}{11}$$

The equation is now solved.

$$x=\frac{2\sqrt{114}+40}{11} x=\frac{40-2\sqrt{114}}{11}$$

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.

$$11x^{2}-80x+104=0$$

Subtract 104 from both sides of the equation.

$$11x^{2}-80x+104-104=-104$$

Subtracting 104 from itself leaves 0.

$$11x^{2}-80x=-104$$

Divide both sides by 11.

$$\frac{11x^{2}-80x}{11}=-\frac{104}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$x^{2}-\frac{80}{11}x=-\frac{104}{11}$$

Divide -\frac{80}{11}, the coefficient of the x term, by 2 to get -\frac{40}{11}. Then add the square of -\frac{40}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{80}{11}x+\left(-\frac{40}{11}\right)^{2}=-\frac{104}{11}+\left(-\frac{40}{11}\right)^{2}$$

Square -\frac{40}{11} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{80}{11}x+\frac{1600}{121}=-\frac{104}{11}+\frac{1600}{121}$$

Add -\frac{104}{11} to \frac{1600}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{80}{11}x+\frac{1600}{121}=\frac{456}{121}$$

Factor x^{2}-\frac{80}{11}x+\frac{1600}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{40}{11}\right)^{2}=\frac{456}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{40}{11}\right)^{2}}=\sqrt{\frac{456}{121}}$$

Simplify.

$$x-\frac{40}{11}=\frac{2\sqrt{114}}{11} x-\frac{40}{11}=-\frac{2\sqrt{114}}{11}$$

Add \frac{40}{11} to both sides of the equation.

$$x=\frac{2\sqrt{114}+40}{11} x=\frac{40-2\sqrt{114}}{11}$$