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7x^{2}+7x+3=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{-7±\sqrt{7^{2}-4\times 7\times 3}}{2\times 7}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 7\times 3}}{2\times 7}
Square 7.
x=\frac{-7±\sqrt{49-28\times 3}}{2\times 7}
Multiply -4 times 7.
x=\frac{-7±\sqrt{49-84}}{2\times 7}
Multiply -28 times 3.
x=\frac{-7±\sqrt{-35}}{2\times 7}
Add 49 to -84.
x=\frac{-7±\sqrt{35}i}{2\times 7}
Take the square root of -35.
x=\frac{-7±\sqrt{35}i}{14}
Multiply 2 times 7.
x=\frac{-7+\sqrt{35}i}{14}
Now solve the equation x=\frac{-7±\sqrt{35}i}{14} when ± is plus. Add -7 to i\sqrt{35}.
x=\frac{\sqrt{35}i}{14}-\frac{1}{2}
Divide -7+i\sqrt{35} by 14.
x=\frac{-\sqrt{35}i-7}{14}
Now solve the equation x=\frac{-7±\sqrt{35}i}{14} when ± is minus. Subtract i\sqrt{35} from -7.
x=-\frac{\sqrt{35}i}{14}-\frac{1}{2}
Divide -7-i\sqrt{35} by 14.
x=\frac{\sqrt{35}i}{14}-\frac{1}{2} x=-\frac{\sqrt{35}i}{14}-\frac{1}{2}
The equation is now solved.
7x^{2}+7x+3=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.
7x^{2}+7x+3-3=-3
Subtract 3 from both sides of the equation.
7x^{2}+7x=-3
Subtracting 3 from itself leaves 0.
\frac{7x^{2}+7x}{7}=-\frac{3}{7}
Divide both sides by 7.
x^{2}+\frac{7}{7}x=-\frac{3}{7}
Dividing by 7 undoes the multiplication by 7.
x^{2}+x=-\frac{3}{7}
Divide 7 by 7.
x^{2}+x+\left(\frac{1}{2}\right)^{2}=-\frac{3}{7}+\left(\frac{1}{2}\right)^{2}
Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+x+\frac{1}{4}=-\frac{3}{7}+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+x+\frac{1}{4}=-\frac{5}{28}
Add -\frac{3}{7} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{2}\right)^{2}=-\frac{5}{28}
Factor x^{2}+x+\frac{1}{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{1}{2}\right)^{2}}=\sqrt{-\frac{5}{28}}
Take the square root of both sides of the equation.
x+\frac{1}{2}=\frac{\sqrt{35}i}{14} x+\frac{1}{2}=-\frac{\sqrt{35}i}{14}
Simplify.
x=\frac{\sqrt{35}i}{14}-\frac{1}{2} x=-\frac{\sqrt{35}i}{14}-\frac{1}{2}
Subtract \frac{1}{2} from both sides of the equation.
x ^ 2 +1x +\frac{3}{7} = 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 7
r + s = -1 rs = \frac{3}{7}
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{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{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{1}{2} - u) (-\frac{1}{2} + u) = \frac{3}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{7}
\frac{1}{4} - u^2 = \frac{3}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{7}-\frac{1}{4} = \frac{5}{28}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{5}{28} u = \pm\sqrt{-\frac{5}{28}} = \pm \frac{\sqrt{5}}{\sqrt{28}}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{2} - \frac{\sqrt{5}}{\sqrt{28}}i = -0.500 - 0.423i s = -\frac{1}{2} + \frac{\sqrt{5}}{\sqrt{28}}i = -0.500 + 0.423i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.