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4x^{2}+24x+2=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{-24±\sqrt{24^{2}-4\times 4\times 2}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 24 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-24±\sqrt{576-4\times 4\times 2}}{2\times 4}
Square 24.
x=\frac{-24±\sqrt{576-16\times 2}}{2\times 4}
Multiply -4 times 4.
x=\frac{-24±\sqrt{576-32}}{2\times 4}
Multiply -16 times 2.
x=\frac{-24±\sqrt{544}}{2\times 4}
Add 576 to -32.
x=\frac{-24±4\sqrt{34}}{2\times 4}
Take the square root of 544.
x=\frac{-24±4\sqrt{34}}{8}
Multiply 2 times 4.
x=\frac{4\sqrt{34}-24}{8}
Now solve the equation x=\frac{-24±4\sqrt{34}}{8} when ± is plus. Add -24 to 4\sqrt{34}.
x=\frac{\sqrt{34}}{2}-3
Divide -24+4\sqrt{34} by 8.
x=\frac{-4\sqrt{34}-24}{8}
Now solve the equation x=\frac{-24±4\sqrt{34}}{8} when ± is minus. Subtract 4\sqrt{34} from -24.
x=-\frac{\sqrt{34}}{2}-3
Divide -24-4\sqrt{34} by 8.
x=\frac{\sqrt{34}}{2}-3 x=-\frac{\sqrt{34}}{2}-3
The equation is now solved.
4x^{2}+24x+2=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.
4x^{2}+24x+2-2=-2
Subtract 2 from both sides of the equation.
4x^{2}+24x=-2
Subtracting 2 from itself leaves 0.
\frac{4x^{2}+24x}{4}=-\frac{2}{4}
Divide both sides by 4.
x^{2}+\frac{24}{4}x=-\frac{2}{4}
Dividing by 4 undoes the multiplication by 4.
x^{2}+6x=-\frac{2}{4}
Divide 24 by 4.
x^{2}+6x=-\frac{1}{2}
Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.
x^{2}+6x+3^{2}=-\frac{1}{2}+3^{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.
x^{2}+6x+9=-\frac{1}{2}+9
Square 3.
x^{2}+6x+9=\frac{17}{2}
Add -\frac{1}{2} to 9.
\left(x+3\right)^{2}=\frac{17}{2}
Factor x^{2}+6x+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(x+3\right)^{2}}=\sqrt{\frac{17}{2}}
Take the square root of both sides of the equation.
x+3=\frac{\sqrt{34}}{2} x+3=-\frac{\sqrt{34}}{2}
Simplify.
x=\frac{\sqrt{34}}{2}-3 x=-\frac{\sqrt{34}}{2}-3
Subtract 3 from both sides of the equation.
x ^ 2 +6x +\frac{1}{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.This is achieved by dividing both sides of the equation by 4
r + s = -6 rs = \frac{1}{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) = \frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}
9 - u^2 = \frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{2}-9 = -\frac{17}{2}
Simplify the expression by subtracting 9 on both sides
u^2 = \frac{17}{2} u = \pm\sqrt{\frac{17}{2}} = \pm \frac{\sqrt{17}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - \frac{\sqrt{17}}{\sqrt{2}} = -5.915 s = -3 + \frac{\sqrt{17}}{\sqrt{2}} = -0.085
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.