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2x^{2}+6x+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{-6±\sqrt{6^{2}-4\times 2\times 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 6 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\times 2\times 2}}{2\times 2}
Square 6.
x=\frac{-6±\sqrt{36-8\times 2}}{2\times 2}
Multiply -4 times 2.
x=\frac{-6±\sqrt{36-16}}{2\times 2}
Multiply -8 times 2.
x=\frac{-6±\sqrt{20}}{2\times 2}
Add 36 to -16.
x=\frac{-6±2\sqrt{5}}{2\times 2}
Take the square root of 20.
x=\frac{-6±2\sqrt{5}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{5}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{5}}{4} when ± is plus. Add -6 to 2\sqrt{5}.
x=\frac{\sqrt{5}-3}{2}
Divide -6+2\sqrt{5} by 4.
x=\frac{-2\sqrt{5}-6}{4}
Now solve the equation x=\frac{-6±2\sqrt{5}}{4} when ± is minus. Subtract 2\sqrt{5} from -6.
x=\frac{-\sqrt{5}-3}{2}
Divide -6-2\sqrt{5} by 4.
x=\frac{\sqrt{5}-3}{2} x=\frac{-\sqrt{5}-3}{2}
The equation is now solved.
2x^{2}+6x+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.
2x^{2}+6x+2-2=-2
Subtract 2 from both sides of the equation.
2x^{2}+6x=-2
Subtracting 2 from itself leaves 0.
\frac{2x^{2}+6x}{2}=-\frac{2}{2}
Divide both sides by 2.
x^{2}+\frac{6}{2}x=-\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+3x=-\frac{2}{2}
Divide 6 by 2.
x^{2}+3x=-1
Divide -2 by 2.
x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-1+\left(\frac{3}{2}\right)^{2}
Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+3x+\frac{9}{4}=-1+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+3x+\frac{9}{4}=\frac{5}{4}
Add -1 to \frac{9}{4}.
\left(x+\frac{3}{2}\right)^{2}=\frac{5}{4}
Factor x^{2}+3x+\frac{9}{4}. 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{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}
Take the square root of both sides of the equation.
x+\frac{3}{2}=\frac{\sqrt{5}}{2} x+\frac{3}{2}=-\frac{\sqrt{5}}{2}
Simplify.
x=\frac{\sqrt{5}-3}{2} x=\frac{-\sqrt{5}-3}{2}
Subtract \frac{3}{2} from both sides of the equation.
x ^ 2 +3x +1 = 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 2
r + s = -3 rs = 1
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{3}{2} - u s = -\frac{3}{2} + u
Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. 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{3}{2} - u) (-\frac{3}{2} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{9}{4} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{9}{4} = -\frac{5}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = \frac{5}{4} u = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{3}{2} - \frac{\sqrt{5}}{2} = -2.618 s = -\frac{3}{2} + \frac{\sqrt{5}}{2} = -0.382
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.