Solve for x
x=\frac{\sqrt{4730}}{33}-\frac{4}{3}\approx 0.750757438
x=-\frac{\sqrt{4730}}{33}-\frac{4}{3}\approx -3.417424105
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99x^{2}+264x-254=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{-264±\sqrt{264^{2}-4\times 99\left(-254\right)}}{2\times 99}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 99 for a, 264 for b, and -254 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-264±\sqrt{69696-4\times 99\left(-254\right)}}{2\times 99}
Square 264.
x=\frac{-264±\sqrt{69696-396\left(-254\right)}}{2\times 99}
Multiply -4 times 99.
x=\frac{-264±\sqrt{69696+100584}}{2\times 99}
Multiply -396 times -254.
x=\frac{-264±\sqrt{170280}}{2\times 99}
Add 69696 to 100584.
x=\frac{-264±6\sqrt{4730}}{2\times 99}
Take the square root of 170280.
x=\frac{-264±6\sqrt{4730}}{198}
Multiply 2 times 99.
x=\frac{6\sqrt{4730}-264}{198}
Now solve the equation x=\frac{-264±6\sqrt{4730}}{198} when ± is plus. Add -264 to 6\sqrt{4730}.
x=\frac{\sqrt{4730}}{33}-\frac{4}{3}
Divide -264+6\sqrt{4730} by 198.
x=\frac{-6\sqrt{4730}-264}{198}
Now solve the equation x=\frac{-264±6\sqrt{4730}}{198} when ± is minus. Subtract 6\sqrt{4730} from -264.
x=-\frac{\sqrt{4730}}{33}-\frac{4}{3}
Divide -264-6\sqrt{4730} by 198.
x=\frac{\sqrt{4730}}{33}-\frac{4}{3} x=-\frac{\sqrt{4730}}{33}-\frac{4}{3}
The equation is now solved.
99x^{2}+264x-254=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.
99x^{2}+264x-254-\left(-254\right)=-\left(-254\right)
Add 254 to both sides of the equation.
99x^{2}+264x=-\left(-254\right)
Subtracting -254 from itself leaves 0.
99x^{2}+264x=254
Subtract -254 from 0.
\frac{99x^{2}+264x}{99}=\frac{254}{99}
Divide both sides by 99.
x^{2}+\frac{264}{99}x=\frac{254}{99}
Dividing by 99 undoes the multiplication by 99.
x^{2}+\frac{8}{3}x=\frac{254}{99}
Reduce the fraction \frac{264}{99} to lowest terms by extracting and canceling out 33.
x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=\frac{254}{99}+\left(\frac{4}{3}\right)^{2}
Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{254}{99}+\frac{16}{9}
Square \frac{4}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{430}{99}
Add \frac{254}{99} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{4}{3}\right)^{2}=\frac{430}{99}
Factor x^{2}+\frac{8}{3}x+\frac{16}{9}. 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{4}{3}\right)^{2}}=\sqrt{\frac{430}{99}}
Take the square root of both sides of the equation.
x+\frac{4}{3}=\frac{\sqrt{4730}}{33} x+\frac{4}{3}=-\frac{\sqrt{4730}}{33}
Simplify.
x=\frac{\sqrt{4730}}{33}-\frac{4}{3} x=-\frac{\sqrt{4730}}{33}-\frac{4}{3}
Subtract \frac{4}{3} from both sides of the equation.
x ^ 2 +\frac{8}{3}x -\frac{254}{99} = 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 99
r + s = -\frac{8}{3} rs = -\frac{254}{99}
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{4}{3} - u s = -\frac{4}{3} + u
Two numbers r and s sum up to -\frac{8}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{8}{3} = -\frac{4}{3}. 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{4}{3} - u) (-\frac{4}{3} + u) = -\frac{254}{99}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{254}{99}
\frac{16}{9} - u^2 = -\frac{254}{99}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{254}{99}-\frac{16}{9} = -\frac{430}{99}
Simplify the expression by subtracting \frac{16}{9} on both sides
u^2 = \frac{430}{99} u = \pm\sqrt{\frac{430}{99}} = \pm \frac{\sqrt{430}}{\sqrt{99}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{4}{3} - \frac{\sqrt{430}}{\sqrt{99}} = -3.417 s = -\frac{4}{3} + \frac{\sqrt{430}}{\sqrt{99}} = 0.751
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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