Solve 400L^2-360L-81=0

$$400 L ^ { 2 } - 360 L - 81 = 0$$

Solve for L

$$L = \frac{9 \sqrt{2} + 9}{20} \approx 1.086396103$$

$$L=\frac{9-9\sqrt{2}}{20}\approx -0.186396103$$

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.

$$400L^{2}-360L-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 400 for a, -360 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$L=\frac{-\left(-360\right)±\sqrt{\left(-360\right)^{2}-4\times 400\left(-81\right)}}{2\times 400}$$

Square -360.

$$L=\frac{-\left(-360\right)±\sqrt{129600-4\times 400\left(-81\right)}}{2\times 400}$$

Multiply -4 times 400.

$$L=\frac{-\left(-360\right)±\sqrt{129600-1600\left(-81\right)}}{2\times 400}$$

Multiply -1600 times -81.

$$L=\frac{-\left(-360\right)±\sqrt{129600+129600}}{2\times 400}$$

Add 129600 to 129600.

$$L=\frac{-\left(-360\right)±\sqrt{259200}}{2\times 400}$$

Take the square root of 259200.

$$L=\frac{-\left(-360\right)±360\sqrt{2}}{2\times 400}$$

The opposite of -360 is 360.

$$L=\frac{360±360\sqrt{2}}{2\times 400}$$

Multiply 2 times 400.

$$L=\frac{360±360\sqrt{2}}{800}$$

Now solve the equation L=\frac{360±360\sqrt{2}}{800} when ± is plus. Add 360 to 360\sqrt{2}.

$$L=\frac{360\sqrt{2}+360}{800}$$

Divide 360+360\sqrt{2} by 800.

$$L=\frac{9\sqrt{2}+9}{20}$$

Now solve the equation L=\frac{360±360\sqrt{2}}{800} when ± is minus. Subtract 360\sqrt{2} from 360.

$$L=\frac{360-360\sqrt{2}}{800}$$

Divide 360-360\sqrt{2} by 800.

$$L=\frac{9-9\sqrt{2}}{20}$$

The equation is now solved.

$$L=\frac{9\sqrt{2}+9}{20} L=\frac{9-9\sqrt{2}}{20}$$

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.

$$400L^{2}-360L-81=0$$

Add 81 to both sides of the equation.

$$400L^{2}-360L-81-\left(-81\right)=-\left(-81\right)$$

Subtracting -81 from itself leaves 0.

$$400L^{2}-360L=-\left(-81\right)$$

Subtract -81 from 0.

$$400L^{2}-360L=81$$

Divide both sides by 400.

$$\frac{400L^{2}-360L}{400}=\frac{81}{400}$$

Dividing by 400 undoes the multiplication by 400.

$$L^{2}+\left(-\frac{360}{400}\right)L=\frac{81}{400}$$

Reduce the fraction \frac{-360}{400} to lowest terms by extracting and canceling out 40.

$$L^{2}-\frac{9}{10}L=\frac{81}{400}$$

Divide -\frac{9}{10}, the coefficient of the x term, by 2 to get -\frac{9}{20}. Then add the square of -\frac{9}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$L^{2}-\frac{9}{10}L+\left(-\frac{9}{20}\right)^{2}=\frac{81}{400}+\left(-\frac{9}{20}\right)^{2}$$

Square -\frac{9}{20} by squaring both the numerator and the denominator of the fraction.

$$L^{2}-\frac{9}{10}L+\frac{81}{400}=\frac{81+81}{400}$$

Add \frac{81}{400} to \frac{81}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$L^{2}-\frac{9}{10}L+\frac{81}{400}=\frac{81}{200}$$

Factor L^{2}-\frac{9}{10}L+\frac{81}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(L-\frac{9}{20}\right)^{2}=\frac{81}{200}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(L-\frac{9}{20}\right)^{2}}=\sqrt{\frac{81}{200}}$$

Simplify.

$$L-\frac{9}{20}=\frac{9\sqrt{2}}{20} L-\frac{9}{20}=-\frac{9\sqrt{2}}{20}$$

Add \frac{9}{20} to both sides of the equation.

$$L=\frac{9\sqrt{2}+9}{20} L=\frac{9-9\sqrt{2}}{20}$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 400

$$x ^ 2 -\frac{9}{10}x -\frac{81}{400} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{9}{10} rs = -\frac{81}{400}$$

Two numbers r and s sum up to \frac{9}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{10} = \frac{9}{20}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{9}{20} - u s = \frac{9}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{81}{400}

$$(\frac{9}{20} - u) (\frac{9}{20} + u) = -\frac{81}{400}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{81}{400} - u^2 = -\frac{81}{400}$$

Simplify the expression by subtracting \frac{81}{400} on both sides

$$-u^2 = -\frac{81}{400}-\frac{81}{400} = -\frac{81}{200}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{81}{200} u = \pm\sqrt{\frac{81}{200}} = \pm \frac{9}{\sqrt{200}}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{9}{20} - \frac{9}{\sqrt{200}} = -0.186 s = \frac{9}{20} + \frac{9}{\sqrt{200}} = 1.086$$