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s^{2}+3s+6=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.
s=\frac{-3±\sqrt{3^{2}-4\times 6}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-3±\sqrt{9-4\times 6}}{2}
Square 3.
s=\frac{-3±\sqrt{9-24}}{2}
Multiply -4 times 6.
s=\frac{-3±\sqrt{-15}}{2}
Add 9 to -24.
s=\frac{-3±\sqrt{15}i}{2}
Take the square root of -15.
s=\frac{-3+\sqrt{15}i}{2}
Now solve the equation s=\frac{-3±\sqrt{15}i}{2} when ± is plus. Add -3 to i\sqrt{15}.
s=\frac{-\sqrt{15}i-3}{2}
Now solve the equation s=\frac{-3±\sqrt{15}i}{2} when ± is minus. Subtract i\sqrt{15} from -3.
s=\frac{-3+\sqrt{15}i}{2} s=\frac{-\sqrt{15}i-3}{2}
The equation is now solved.
s^{2}+3s+6=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.
s^{2}+3s+6-6=-6
Subtract 6 from both sides of the equation.
s^{2}+3s=-6
Subtracting 6 from itself leaves 0.
s^{2}+3s+\left(\frac{3}{2}\right)^{2}=-6+\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.
s^{2}+3s+\frac{9}{4}=-6+\frac{9}{4}
Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.
s^{2}+3s+\frac{9}{4}=-\frac{15}{4}
Add -6 to \frac{9}{4}.
\left(s+\frac{3}{2}\right)^{2}=-\frac{15}{4}
Factor s^{2}+3s+\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(s+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{15}{4}}
Take the square root of both sides of the equation.
s+\frac{3}{2}=\frac{\sqrt{15}i}{2} s+\frac{3}{2}=-\frac{\sqrt{15}i}{2}
Simplify.
s=\frac{-3+\sqrt{15}i}{2} s=\frac{-\sqrt{15}i-3}{2}
Subtract \frac{3}{2} from both sides of the equation.
x ^ 2 +3x +6 = 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 = -3 rs = 6
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) = 6
To solve for unknown quantity u, substitute these in the product equation rs = 6
\frac{9}{4} - u^2 = 6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 6-\frac{9}{4} = \frac{15}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = -\frac{15}{4} u = \pm\sqrt{-\frac{15}{4}} = \pm \frac{\sqrt{15}}{2}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{3}{2} - \frac{\sqrt{15}}{2}i = -1.500 - 1.936i s = -\frac{3}{2} + \frac{\sqrt{15}}{2}i = -1.500 + 1.936i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.