Solve for x
x=\frac{\sqrt{3197}+1}{102}\approx 0.564137449
x=\frac{1-\sqrt{3197}}{102}\approx -0.544529606
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1530x^{2}-30x-470=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.
x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 1530\left(-470\right)}}{2\times 1530}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1530 for a, -30 for b, and -470 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-30\right)±\sqrt{900-4\times 1530\left(-470\right)}}{2\times 1530}
Square -30.
x=\frac{-\left(-30\right)±\sqrt{900-6120\left(-470\right)}}{2\times 1530}
Multiply -4 times 1530.
x=\frac{-\left(-30\right)±\sqrt{900+2876400}}{2\times 1530}
Multiply -6120 times -470.
x=\frac{-\left(-30\right)±\sqrt{2877300}}{2\times 1530}
Add 900 to 2876400.
x=\frac{-\left(-30\right)±30\sqrt{3197}}{2\times 1530}
Take the square root of 2877300.
x=\frac{30±30\sqrt{3197}}{2\times 1530}
The opposite of -30 is 30.
x=\frac{30±30\sqrt{3197}}{3060}
Multiply 2 times 1530.
x=\frac{30\sqrt{3197}+30}{3060}
Now solve the equation x=\frac{30±30\sqrt{3197}}{3060} when ± is plus. Add 30 to 30\sqrt{3197}.
x=\frac{\sqrt{3197}+1}{102}
Divide 30+30\sqrt{3197} by 3060.
x=\frac{30-30\sqrt{3197}}{3060}
Now solve the equation x=\frac{30±30\sqrt{3197}}{3060} when ± is minus. Subtract 30\sqrt{3197} from 30.
x=\frac{1-\sqrt{3197}}{102}
Divide 30-30\sqrt{3197} by 3060.
x=\frac{\sqrt{3197}+1}{102} x=\frac{1-\sqrt{3197}}{102}
The equation is now solved.
1530x^{2}-30x-470=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.
1530x^{2}-30x-470-\left(-470\right)=-\left(-470\right)
Add 470 to both sides of the equation.
1530x^{2}-30x=-\left(-470\right)
Subtracting -470 from itself leaves 0.
1530x^{2}-30x=470
Subtract -470 from 0.
\frac{1530x^{2}-30x}{1530}=\frac{470}{1530}
Divide both sides by 1530.
x^{2}+\left(-\frac{30}{1530}\right)x=\frac{470}{1530}
Dividing by 1530 undoes the multiplication by 1530.
x^{2}-\frac{1}{51}x=\frac{470}{1530}
Reduce the fraction \frac{-30}{1530} to lowest terms by extracting and canceling out 30.
x^{2}-\frac{1}{51}x=\frac{47}{153}
Reduce the fraction \frac{470}{1530} to lowest terms by extracting and canceling out 10.
x^{2}-\frac{1}{51}x+\left(-\frac{1}{102}\right)^{2}=\frac{47}{153}+\left(-\frac{1}{102}\right)^{2}
Divide -\frac{1}{51}, the coefficient of the x term, by 2 to get -\frac{1}{102}. Then add the square of -\frac{1}{102} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{51}x+\frac{1}{10404}=\frac{47}{153}+\frac{1}{10404}
Square -\frac{1}{102} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{51}x+\frac{1}{10404}=\frac{3197}{10404}
Add \frac{47}{153} to \frac{1}{10404} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{102}\right)^{2}=\frac{3197}{10404}
Factor x^{2}-\frac{1}{51}x+\frac{1}{10404}. 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(x-\frac{1}{102}\right)^{2}}=\sqrt{\frac{3197}{10404}}
Take the square root of both sides of the equation.
x-\frac{1}{102}=\frac{\sqrt{3197}}{102} x-\frac{1}{102}=-\frac{\sqrt{3197}}{102}
Simplify.
x=\frac{\sqrt{3197}+1}{102} x=\frac{1-\sqrt{3197}}{102}
Add \frac{1}{102} to both sides of the equation.
x ^ 2 -\frac{1}{51}x -\frac{47}{153} = 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 1530
r + s = \frac{1}{51} rs = -\frac{47}{153}
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{1}{102} - u s = \frac{1}{102} + u
Two numbers r and s sum up to \frac{1}{51} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{51} = \frac{1}{102}. 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{1}{102} - u) (\frac{1}{102} + u) = -\frac{47}{153}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{47}{153}
\frac{1}{10404} - u^2 = -\frac{47}{153}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{47}{153}-\frac{1}{10404} = -\frac{3197}{10404}
Simplify the expression by subtracting \frac{1}{10404} on both sides
u^2 = \frac{3197}{10404} u = \pm\sqrt{\frac{3197}{10404}} = \pm \frac{\sqrt{3197}}{102}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{102} - \frac{\sqrt{3197}}{102} = -0.545 s = \frac{1}{102} + \frac{\sqrt{3197}}{102} = 0.564
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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