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-2x^{2}-16x-44=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(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-2\right)\left(-44\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -16 for b, and -44 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-16\right)±\sqrt{256-4\left(-2\right)\left(-44\right)}}{2\left(-2\right)}
Square -16.
x=\frac{-\left(-16\right)±\sqrt{256+8\left(-44\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-\left(-16\right)±\sqrt{256-352}}{2\left(-2\right)}
Multiply 8 times -44.
x=\frac{-\left(-16\right)±\sqrt{-96}}{2\left(-2\right)}
Add 256 to -352.
x=\frac{-\left(-16\right)±4\sqrt{6}i}{2\left(-2\right)}
Take the square root of -96.
x=\frac{16±4\sqrt{6}i}{2\left(-2\right)}
The opposite of -16 is 16.
x=\frac{16±4\sqrt{6}i}{-4}
Multiply 2 times -2.
x=\frac{16+4\sqrt{6}i}{-4}
Now solve the equation x=\frac{16±4\sqrt{6}i}{-4} when ± is plus. Add 16 to 4i\sqrt{6}.
x=-\sqrt{6}i-4
Divide 16+4i\sqrt{6} by -4.
x=\frac{-4\sqrt{6}i+16}{-4}
Now solve the equation x=\frac{16±4\sqrt{6}i}{-4} when ± is minus. Subtract 4i\sqrt{6} from 16.
x=-4+\sqrt{6}i
Divide 16-4i\sqrt{6} by -4.
x=-\sqrt{6}i-4 x=-4+\sqrt{6}i
The equation is now solved.
-2x^{2}-16x-44=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.
-2x^{2}-16x-44-\left(-44\right)=-\left(-44\right)
Add 44 to both sides of the equation.
-2x^{2}-16x=-\left(-44\right)
Subtracting -44 from itself leaves 0.
-2x^{2}-16x=44
Subtract -44 from 0.
\frac{-2x^{2}-16x}{-2}=\frac{44}{-2}
Divide both sides by -2.
x^{2}+\left(-\frac{16}{-2}\right)x=\frac{44}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}+8x=\frac{44}{-2}
Divide -16 by -2.
x^{2}+8x=-22
Divide 44 by -2.
x^{2}+8x+4^{2}=-22+4^{2}
Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+8x+16=-22+16
Square 4.
x^{2}+8x+16=-6
Add -22 to 16.
\left(x+4\right)^{2}=-6
Factor x^{2}+8x+16. 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+4\right)^{2}}=\sqrt{-6}
Take the square root of both sides of the equation.
x+4=\sqrt{6}i x+4=-\sqrt{6}i
Simplify.
x=-4+\sqrt{6}i x=-\sqrt{6}i-4
Subtract 4 from both sides of the equation.
x ^ 2 +8x +22 = 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 = -8 rs = 22
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 = -4 - u s = -4 + u
Two numbers r and s sum up to -8 exactly when the average of the two numbers is \frac{1}{2}*-8 = -4. 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>
(-4 - u) (-4 + u) = 22
To solve for unknown quantity u, substitute these in the product equation rs = 22
16 - u^2 = 22
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 22-16 = 6
Simplify the expression by subtracting 16 on both sides
u^2 = -6 u = \pm\sqrt{-6} = \pm \sqrt{6}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-4 - \sqrt{6}i s = -4 + \sqrt{6}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.