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-x^{2}+6x+15=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\left(-1\right)\times 15}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\left(-1\right)\times 15}}{2\left(-1\right)}
Square 6.
x=\frac{-6±\sqrt{36+4\times 15}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-6±\sqrt{36+60}}{2\left(-1\right)}
Multiply 4 times 15.
x=\frac{-6±\sqrt{96}}{2\left(-1\right)}
Add 36 to 60.
x=\frac{-6±4\sqrt{6}}{2\left(-1\right)}
Take the square root of 96.
x=\frac{-6±4\sqrt{6}}{-2}
Multiply 2 times -1.
x=\frac{4\sqrt{6}-6}{-2}
Now solve the equation x=\frac{-6±4\sqrt{6}}{-2} when ± is plus. Add -6 to 4\sqrt{6}.
x=3-2\sqrt{6}
Divide -6+4\sqrt{6} by -2.
x=\frac{-4\sqrt{6}-6}{-2}
Now solve the equation x=\frac{-6±4\sqrt{6}}{-2} when ± is minus. Subtract 4\sqrt{6} from -6.
x=2\sqrt{6}+3
Divide -6-4\sqrt{6} by -2.
x=3-2\sqrt{6} x=2\sqrt{6}+3
The equation is now solved.
-x^{2}+6x+15=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.
-x^{2}+6x+15-15=-15
Subtract 15 from both sides of the equation.
-x^{2}+6x=-15
Subtracting 15 from itself leaves 0.
\frac{-x^{2}+6x}{-1}=-\frac{15}{-1}
Divide both sides by -1.
x^{2}+\frac{6}{-1}x=-\frac{15}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}-6x=-\frac{15}{-1}
Divide 6 by -1.
x^{2}-6x=15
Divide -15 by -1.
x^{2}-6x+\left(-3\right)^{2}=15+\left(-3\right)^{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.
x^{2}-6x+9=15+9
Square -3.
x^{2}-6x+9=24
Add 15 to 9.
\left(x-3\right)^{2}=24
Factor x^{2}-6x+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(x-3\right)^{2}}=\sqrt{24}
Take the square root of both sides of the equation.
x-3=2\sqrt{6} x-3=-2\sqrt{6}
Simplify.
x=2\sqrt{6}+3 x=3-2\sqrt{6}
Add 3 to both sides of the equation.
x ^ 2 -6x -15 = 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 = -15
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) = -15
To solve for unknown quantity u, substitute these in the product equation rs = -15
9 - u^2 = -15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -15-9 = -24
Simplify the expression by subtracting 9 on both sides
u^2 = 24 u = \pm\sqrt{24} = \pm \sqrt{24}
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{24} = -1.899 s = 3 + \sqrt{24} = 7.899
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.