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-9x^{2}-4x-6=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(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-9\right)\left(-6\right)}}{2\left(-9\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\left(-9\right)\left(-6\right)}}{2\left(-9\right)}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16+36\left(-6\right)}}{2\left(-9\right)}
Multiply -4 times -9.
x=\frac{-\left(-4\right)±\sqrt{16-216}}{2\left(-9\right)}
Multiply 36 times -6.
x=\frac{-\left(-4\right)±\sqrt{-200}}{2\left(-9\right)}
Add 16 to -216.
x=\frac{-\left(-4\right)±10\sqrt{2}i}{2\left(-9\right)}
Take the square root of -200.
x=\frac{4±10\sqrt{2}i}{2\left(-9\right)}
The opposite of -4 is 4.
x=\frac{4±10\sqrt{2}i}{-18}
Multiply 2 times -9.
x=\frac{4+10\sqrt{2}i}{-18}
Now solve the equation x=\frac{4±10\sqrt{2}i}{-18} when ± is plus. Add 4 to 10i\sqrt{2}.
x=\frac{-5\sqrt{2}i-2}{9}
Divide 4+10i\sqrt{2} by -18.
x=\frac{-10\sqrt{2}i+4}{-18}
Now solve the equation x=\frac{4±10\sqrt{2}i}{-18} when ± is minus. Subtract 10i\sqrt{2} from 4.
x=\frac{-2+5\sqrt{2}i}{9}
Divide 4-10i\sqrt{2} by -18.
x=\frac{-5\sqrt{2}i-2}{9} x=\frac{-2+5\sqrt{2}i}{9}
The equation is now solved.
-9x^{2}-4x-6=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.
-9x^{2}-4x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
-9x^{2}-4x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
-9x^{2}-4x=6
Subtract -6 from 0.
\frac{-9x^{2}-4x}{-9}=\frac{6}{-9}
Divide both sides by -9.
x^{2}+\left(-\frac{4}{-9}\right)x=\frac{6}{-9}
Dividing by -9 undoes the multiplication by -9.
x^{2}+\frac{4}{9}x=\frac{6}{-9}
Divide -4 by -9.
x^{2}+\frac{4}{9}x=-\frac{2}{3}
Reduce the fraction \frac{6}{-9} to lowest terms by extracting and canceling out 3.
x^{2}+\frac{4}{9}x+\left(\frac{2}{9}\right)^{2}=-\frac{2}{3}+\left(\frac{2}{9}\right)^{2}
Divide \frac{4}{9}, the coefficient of the x term, by 2 to get \frac{2}{9}. Then add the square of \frac{2}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{4}{9}x+\frac{4}{81}=-\frac{2}{3}+\frac{4}{81}
Square \frac{2}{9} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{4}{9}x+\frac{4}{81}=-\frac{50}{81}
Add -\frac{2}{3} to \frac{4}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{2}{9}\right)^{2}=-\frac{50}{81}
Factor x^{2}+\frac{4}{9}x+\frac{4}{81}. 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{2}{9}\right)^{2}}=\sqrt{-\frac{50}{81}}
Take the square root of both sides of the equation.
x+\frac{2}{9}=\frac{5\sqrt{2}i}{9} x+\frac{2}{9}=-\frac{5\sqrt{2}i}{9}
Simplify.
x=\frac{-2+5\sqrt{2}i}{9} x=\frac{-5\sqrt{2}i-2}{9}
Subtract \frac{2}{9} from both sides of the equation.
x ^ 2 +\frac{4}{9}x +\frac{2}{3} = 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.
r + s = -\frac{4}{9} rs = \frac{2}{3}
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{2}{9} - u s = -\frac{2}{9} + u
Two numbers r and s sum up to -\frac{4}{9} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{9} = -\frac{2}{9}. 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{2}{9} - u) (-\frac{2}{9} + u) = \frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}
\frac{4}{81} - u^2 = \frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{3}-\frac{4}{81} = \frac{50}{81}
Simplify the expression by subtracting \frac{4}{81} on both sides
u^2 = -\frac{50}{81} u = \pm\sqrt{-\frac{50}{81}} = \pm \frac{\sqrt{50}}{9}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{2}{9} - \frac{\sqrt{50}}{9}i = -0.222 - 0.786i s = -\frac{2}{9} + \frac{\sqrt{50}}{9}i = -0.222 + 0.786i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.