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100a^{2}-60a-60=0
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.
a=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 100\left(-60\right)}}{2\times 100}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, -60 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
a=\frac{-\left(-60\right)±\sqrt{3600-4\times 100\left(-60\right)}}{2\times 100}
Square -60.
a=\frac{-\left(-60\right)±\sqrt{3600-400\left(-60\right)}}{2\times 100}
Multiply -4 times 100.
a=\frac{-\left(-60\right)±\sqrt{3600+24000}}{2\times 100}
Multiply -400 times -60.
a=\frac{-\left(-60\right)±\sqrt{27600}}{2\times 100}
Add 3600 to 24000.
a=\frac{-\left(-60\right)±20\sqrt{69}}{2\times 100}
Take the square root of 27600.
a=\frac{60±20\sqrt{69}}{2\times 100}
The opposite of -60 is 60.
a=\frac{60±20\sqrt{69}}{200}
Multiply 2 times 100.
a=\frac{20\sqrt{69}+60}{200}
Now solve the equation a=\frac{60±20\sqrt{69}}{200} when ± is plus. Add 60 to 20\sqrt{69}.
a=\frac{\sqrt{69}+3}{10}
Divide 60+20\sqrt{69} by 200.
a=\frac{60-20\sqrt{69}}{200}
Now solve the equation a=\frac{60±20\sqrt{69}}{200} when ± is minus. Subtract 20\sqrt{69} from 60.
a=\frac{3-\sqrt{69}}{10}
Divide 60-20\sqrt{69} by 200.
a=\frac{\sqrt{69}+3}{10} a=\frac{3-\sqrt{69}}{10}
The equation is now solved.
100a^{2}-60a-60=0
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.
100a^{2}-60a-60-\left(-60\right)=-\left(-60\right)
Add 60 to both sides of the equation.
100a^{2}-60a=-\left(-60\right)
Subtracting -60 from itself leaves 0.
100a^{2}-60a=60
Subtract -60 from 0.
\frac{100a^{2}-60a}{100}=\frac{60}{100}
Divide both sides by 100.
a^{2}+\left(-\frac{60}{100}\right)a=\frac{60}{100}
Dividing by 100 undoes the multiplication by 100.
a^{2}-\frac{3}{5}a=\frac{60}{100}
Reduce the fraction \frac{-60}{100} to lowest terms by extracting and canceling out 20.
a^{2}-\frac{3}{5}a=\frac{3}{5}
Reduce the fraction \frac{60}{100} to lowest terms by extracting and canceling out 20.
a^{2}-\frac{3}{5}a+\left(-\frac{3}{10}\right)^{2}=\frac{3}{5}+\left(-\frac{3}{10}\right)^{2}
Divide -\frac{3}{5}, the coefficient of the x term, by 2 to get -\frac{3}{10}. Then add the square of -\frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
a^{2}-\frac{3}{5}a+\frac{9}{100}=\frac{3}{5}+\frac{9}{100}
Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.
a^{2}-\frac{3}{5}a+\frac{9}{100}=\frac{69}{100}
Add \frac{3}{5} to \frac{9}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(a-\frac{3}{10}\right)^{2}=\frac{69}{100}
Factor a^{2}-\frac{3}{5}a+\frac{9}{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}.
\sqrt{\left(a-\frac{3}{10}\right)^{2}}=\sqrt{\frac{69}{100}}
Take the square root of both sides of the equation.
a-\frac{3}{10}=\frac{\sqrt{69}}{10} a-\frac{3}{10}=-\frac{\sqrt{69}}{10}
Simplify.
a=\frac{\sqrt{69}+3}{10} a=\frac{3-\sqrt{69}}{10}
Add \frac{3}{10} to both sides of the equation.
x ^ 2 -\frac{3}{5}x -\frac{3}{5} = 0
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 100
r + s = \frac{3}{5} rs = -\frac{3}{5}
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 = \frac{3}{10} - u s = \frac{3}{10} + u
Two numbers r and s sum up to \frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{5} = \frac{3}{10}. 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>
(\frac{3}{10} - u) (\frac{3}{10} + u) = -\frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}
\frac{9}{100} - u^2 = -\frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{5}-\frac{9}{100} = -\frac{69}{100}
Simplify the expression by subtracting \frac{9}{100} on both sides
u^2 = \frac{69}{100} u = \pm\sqrt{\frac{69}{100}} = \pm \frac{\sqrt{69}}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{10} - \frac{\sqrt{69}}{10} = -0.531 s = \frac{3}{10} + \frac{\sqrt{69}}{10} = 1.131
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.