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-2z^{2}+12z-4=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.
z=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-4\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-12±\sqrt{144-4\left(-2\right)\left(-4\right)}}{2\left(-2\right)}
Square 12.
z=\frac{-12±\sqrt{144+8\left(-4\right)}}{2\left(-2\right)}
Multiply -4 times -2.
z=\frac{-12±\sqrt{144-32}}{2\left(-2\right)}
Multiply 8 times -4.
z=\frac{-12±\sqrt{112}}{2\left(-2\right)}
Add 144 to -32.
z=\frac{-12±4\sqrt{7}}{2\left(-2\right)}
Take the square root of 112.
z=\frac{-12±4\sqrt{7}}{-4}
Multiply 2 times -2.
z=\frac{4\sqrt{7}-12}{-4}
Now solve the equation z=\frac{-12±4\sqrt{7}}{-4} when ± is plus. Add -12 to 4\sqrt{7}.
z=3-\sqrt{7}
Divide -12+4\sqrt{7} by -4.
z=\frac{-4\sqrt{7}-12}{-4}
Now solve the equation z=\frac{-12±4\sqrt{7}}{-4} when ± is minus. Subtract 4\sqrt{7} from -12.
z=\sqrt{7}+3
Divide -12-4\sqrt{7} by -4.
z=3-\sqrt{7} z=\sqrt{7}+3
The equation is now solved.
-2z^{2}+12z-4=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.
-2z^{2}+12z-4-\left(-4\right)=-\left(-4\right)
Add 4 to both sides of the equation.
-2z^{2}+12z=-\left(-4\right)
Subtracting -4 from itself leaves 0.
-2z^{2}+12z=4
Subtract -4 from 0.
\frac{-2z^{2}+12z}{-2}=\frac{4}{-2}
Divide both sides by -2.
z^{2}+\frac{12}{-2}z=\frac{4}{-2}
Dividing by -2 undoes the multiplication by -2.
z^{2}-6z=\frac{4}{-2}
Divide 12 by -2.
z^{2}-6z=-2
Divide 4 by -2.
z^{2}-6z+\left(-3\right)^{2}=-2+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
z^{2}-6z+9=-2+9
Square -3.
z^{2}-6z+9=7
Add -2 to 9.
\left(z-3\right)^{2}=7
Factor z^{2}-6z+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(z-3\right)^{2}}=\sqrt{7}
Take the square root of both sides of the equation.
z-3=\sqrt{7} z-3=-\sqrt{7}
Simplify.
z=\sqrt{7}+3 z=3-\sqrt{7}
Add 3 to both sides of the equation.
x ^ 2 -6x +2 = 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 = 6 rs = 2
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 = 3 - u s = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 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>
(3 - u) (3 + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
9 - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-9 = -7
Simplify the expression by subtracting 9 on both sides
u^2 = 7 u = \pm\sqrt{7} = \pm \sqrt{7}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - \sqrt{7} = 0.354 s = 3 + \sqrt{7} = 5.646
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.