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3v^{2}+9v+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.
v=\frac{-9±\sqrt{9^{2}-4\times 3\times 3}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 9 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
v=\frac{-9±\sqrt{81-4\times 3\times 3}}{2\times 3}
Square 9.
v=\frac{-9±\sqrt{81-12\times 3}}{2\times 3}
Multiply -4 times 3.
v=\frac{-9±\sqrt{81-36}}{2\times 3}
Multiply -12 times 3.
v=\frac{-9±\sqrt{45}}{2\times 3}
Add 81 to -36.
v=\frac{-9±3\sqrt{5}}{2\times 3}
Take the square root of 45.
v=\frac{-9±3\sqrt{5}}{6}
Multiply 2 times 3.
v=\frac{3\sqrt{5}-9}{6}
Now solve the equation v=\frac{-9±3\sqrt{5}}{6} when ± is plus. Add -9 to 3\sqrt{5}.
v=\frac{\sqrt{5}-3}{2}
Divide -9+3\sqrt{5} by 6.
v=\frac{-3\sqrt{5}-9}{6}
Now solve the equation v=\frac{-9±3\sqrt{5}}{6} when ± is minus. Subtract 3\sqrt{5} from -9.
v=\frac{-\sqrt{5}-3}{2}
Divide -9-3\sqrt{5} by 6.
v=\frac{\sqrt{5}-3}{2} v=\frac{-\sqrt{5}-3}{2}
The equation is now solved.
3v^{2}+9v+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.
3v^{2}+9v+3-3=-3
Subtract 3 from both sides of the equation.
3v^{2}+9v=-3
Subtracting 3 from itself leaves 0.
\frac{3v^{2}+9v}{3}=-\frac{3}{3}
Divide both sides by 3.
v^{2}+\frac{9}{3}v=-\frac{3}{3}
Dividing by 3 undoes the multiplication by 3.
v^{2}+3v=-\frac{3}{3}
Divide 9 by 3.
v^{2}+3v=-1
Divide -3 by 3.
v^{2}+3v+\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.
v^{2}+3v+\frac{9}{4}=-1+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
v^{2}+3v+\frac{9}{4}=\frac{5}{4}
Add -1 to \frac{9}{4}.
\left(v+\frac{3}{2}\right)^{2}=\frac{5}{4}
Factor v^{2}+3v+\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(v+\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}
Take the square root of both sides of the equation.
v+\frac{3}{2}=\frac{\sqrt{5}}{2} v+\frac{3}{2}=-\frac{\sqrt{5}}{2}
Simplify.
v=\frac{\sqrt{5}-3}{2} v=\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 3
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.