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-3x^{2}-6x-9=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(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-6\right)±\sqrt{36-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36+12\left(-9\right)}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-\left(-6\right)±\sqrt{36-108}}{2\left(-3\right)}
Multiply 12 times -9.
x=\frac{-\left(-6\right)±\sqrt{-72}}{2\left(-3\right)}
Add 36 to -108.
x=\frac{-\left(-6\right)±6\sqrt{2}i}{2\left(-3\right)}
Take the square root of -72.
x=\frac{6±6\sqrt{2}i}{2\left(-3\right)}
The opposite of -6 is 6.
x=\frac{6±6\sqrt{2}i}{-6}
Multiply 2 times -3.
x=\frac{6+6\sqrt{2}i}{-6}
Now solve the equation x=\frac{6±6\sqrt{2}i}{-6} when ± is plus. Add 6 to 6i\sqrt{2}.
x=-\sqrt{2}i-1
Divide 6+6i\sqrt{2} by -6.
x=\frac{-6\sqrt{2}i+6}{-6}
Now solve the equation x=\frac{6±6\sqrt{2}i}{-6} when ± is minus. Subtract 6i\sqrt{2} from 6.
x=-1+\sqrt{2}i
Divide 6-6i\sqrt{2} by -6.
x=-\sqrt{2}i-1 x=-1+\sqrt{2}i
The equation is now solved.
-3x^{2}-6x-9=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.
-3x^{2}-6x-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
-3x^{2}-6x=-\left(-9\right)
Subtracting -9 from itself leaves 0.
-3x^{2}-6x=9
Subtract -9 from 0.
\frac{-3x^{2}-6x}{-3}=\frac{9}{-3}
Divide both sides by -3.
x^{2}+\left(-\frac{6}{-3}\right)x=\frac{9}{-3}
Dividing by -3 undoes the multiplication by -3.
x^{2}+2x=\frac{9}{-3}
Divide -6 by -3.
x^{2}+2x=-3
Divide 9 by -3.
x^{2}+2x+1^{2}=-3+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=-3+1
Square 1.
x^{2}+2x+1=-2
Add -3 to 1.
\left(x+1\right)^{2}=-2
Factor x^{2}+2x+1. 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+1\right)^{2}}=\sqrt{-2}
Take the square root of both sides of the equation.
x+1=\sqrt{2}i x+1=-\sqrt{2}i
Simplify.
x=-1+\sqrt{2}i x=-\sqrt{2}i-1
Subtract 1 from both sides of the equation.
x ^ 2 +2x +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 = -2 rs = 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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = 3
To solve for unknown quantity u, substitute these in the product equation rs = 3
1 - u^2 = 3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 3-1 = 2
Simplify the expression by subtracting 1 on both sides
u^2 = -2 u = \pm\sqrt{-2} = \pm \sqrt{2}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \sqrt{2}i s = -1 + \sqrt{2}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.