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33d^{2}+22d-36=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.
d=\frac{-22±\sqrt{22^{2}-4\times 33\left(-36\right)}}{2\times 33}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 33 for a, 22 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
d=\frac{-22±\sqrt{484-4\times 33\left(-36\right)}}{2\times 33}
Square 22.
d=\frac{-22±\sqrt{484-132\left(-36\right)}}{2\times 33}
Multiply -4 times 33.
d=\frac{-22±\sqrt{484+4752}}{2\times 33}
Multiply -132 times -36.
d=\frac{-22±\sqrt{5236}}{2\times 33}
Add 484 to 4752.
d=\frac{-22±2\sqrt{1309}}{2\times 33}
Take the square root of 5236.
d=\frac{-22±2\sqrt{1309}}{66}
Multiply 2 times 33.
d=\frac{2\sqrt{1309}-22}{66}
Now solve the equation d=\frac{-22±2\sqrt{1309}}{66} when ± is plus. Add -22 to 2\sqrt{1309}.
d=\frac{\sqrt{1309}}{33}-\frac{1}{3}
Divide -22+2\sqrt{1309} by 66.
d=\frac{-2\sqrt{1309}-22}{66}
Now solve the equation d=\frac{-22±2\sqrt{1309}}{66} when ± is minus. Subtract 2\sqrt{1309} from -22.
d=-\frac{\sqrt{1309}}{33}-\frac{1}{3}
Divide -22-2\sqrt{1309} by 66.
d=\frac{\sqrt{1309}}{33}-\frac{1}{3} d=-\frac{\sqrt{1309}}{33}-\frac{1}{3}
The equation is now solved.
33d^{2}+22d-36=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.
33d^{2}+22d-36-\left(-36\right)=-\left(-36\right)
Add 36 to both sides of the equation.
33d^{2}+22d=-\left(-36\right)
Subtracting -36 from itself leaves 0.
33d^{2}+22d=36
Subtract -36 from 0.
\frac{33d^{2}+22d}{33}=\frac{36}{33}
Divide both sides by 33.
d^{2}+\frac{22}{33}d=\frac{36}{33}
Dividing by 33 undoes the multiplication by 33.
d^{2}+\frac{2}{3}d=\frac{36}{33}
Reduce the fraction \frac{22}{33} to lowest terms by extracting and canceling out 11.
d^{2}+\frac{2}{3}d=\frac{12}{11}
Reduce the fraction \frac{36}{33} to lowest terms by extracting and canceling out 3.
d^{2}+\frac{2}{3}d+\left(\frac{1}{3}\right)^{2}=\frac{12}{11}+\left(\frac{1}{3}\right)^{2}
Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
d^{2}+\frac{2}{3}d+\frac{1}{9}=\frac{12}{11}+\frac{1}{9}
Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.
d^{2}+\frac{2}{3}d+\frac{1}{9}=\frac{119}{99}
Add \frac{12}{11} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(d+\frac{1}{3}\right)^{2}=\frac{119}{99}
Factor d^{2}+\frac{2}{3}d+\frac{1}{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(d+\frac{1}{3}\right)^{2}}=\sqrt{\frac{119}{99}}
Take the square root of both sides of the equation.
d+\frac{1}{3}=\frac{\sqrt{1309}}{33} d+\frac{1}{3}=-\frac{\sqrt{1309}}{33}
Simplify.
d=\frac{\sqrt{1309}}{33}-\frac{1}{3} d=-\frac{\sqrt{1309}}{33}-\frac{1}{3}
Subtract \frac{1}{3} from both sides of the equation.
x ^ 2 +\frac{2}{3}x -\frac{12}{11} = 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 33
r + s = -\frac{2}{3} rs = -\frac{12}{11}
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}{3} - u s = -\frac{1}{3} + u
Two numbers r and s sum up to -\frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{3} = -\frac{1}{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{1}{3} - u) (-\frac{1}{3} + u) = -\frac{12}{11}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{12}{11}
\frac{1}{9} - u^2 = -\frac{12}{11}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{12}{11}-\frac{1}{9} = -\frac{119}{99}
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = \frac{119}{99} u = \pm\sqrt{\frac{119}{99}} = \pm \frac{\sqrt{119}}{\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{1}{3} - \frac{\sqrt{119}}{\sqrt{99}} = -1.430 s = -\frac{1}{3} + \frac{\sqrt{119}}{\sqrt{99}} = 0.763
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.