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y^{2}+6y-9=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.
y=\frac{-6±\sqrt{6^{2}-4\left(-9\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-6±\sqrt{36-4\left(-9\right)}}{2}
Square 6.
y=\frac{-6±\sqrt{36+36}}{2}
Multiply -4 times -9.
y=\frac{-6±\sqrt{72}}{2}
Add 36 to 36.
y=\frac{-6±6\sqrt{2}}{2}
Take the square root of 72.
y=\frac{6\sqrt{2}-6}{2}
Now solve the equation y=\frac{-6±6\sqrt{2}}{2} when ± is plus. Add -6 to 6\sqrt{2}.
y=3\sqrt{2}-3
Divide -6+6\sqrt{2} by 2.
y=\frac{-6\sqrt{2}-6}{2}
Now solve the equation y=\frac{-6±6\sqrt{2}}{2} when ± is minus. Subtract 6\sqrt{2} from -6.
y=-3\sqrt{2}-3
Divide -6-6\sqrt{2} by 2.
y=3\sqrt{2}-3 y=-3\sqrt{2}-3
The equation is now solved.
y^{2}+6y-9=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.
y^{2}+6y-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
y^{2}+6y=-\left(-9\right)
Subtracting -9 from itself leaves 0.
y^{2}+6y=9
Subtract -9 from 0.
y^{2}+6y+3^{2}=9+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.
y^{2}+6y+9=9+9
Square 3.
y^{2}+6y+9=18
Add 9 to 9.
\left(y+3\right)^{2}=18
Factor y^{2}+6y+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(y+3\right)^{2}}=\sqrt{18}
Take the square root of both sides of the equation.
y+3=3\sqrt{2} y+3=-3\sqrt{2}
Simplify.
y=3\sqrt{2}-3 y=-3\sqrt{2}-3
Subtract 3 from both sides of the equation.
x ^ 2 +6x -9 = 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 = -6 rs = -9
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) = -9
To solve for unknown quantity u, substitute these in the product equation rs = -9
9 - u^2 = -9
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -9-9 = -18
Simplify the expression by subtracting 9 on both sides
u^2 = 18 u = \pm\sqrt{18} = \pm \sqrt{18}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - \sqrt{18} = -7.243 s = -3 + \sqrt{18} = 1.243
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.