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-3k^{2}-18k+57=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.
k=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-3\right)\times 57}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -18 for b, and 57 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
k=\frac{-\left(-18\right)±\sqrt{324-4\left(-3\right)\times 57}}{2\left(-3\right)}
Square -18.
k=\frac{-\left(-18\right)±\sqrt{324+12\times 57}}{2\left(-3\right)}
Multiply -4 times -3.
k=\frac{-\left(-18\right)±\sqrt{324+684}}{2\left(-3\right)}
Multiply 12 times 57.
k=\frac{-\left(-18\right)±\sqrt{1008}}{2\left(-3\right)}
Add 324 to 684.
k=\frac{-\left(-18\right)±12\sqrt{7}}{2\left(-3\right)}
Take the square root of 1008.
k=\frac{18±12\sqrt{7}}{2\left(-3\right)}
The opposite of -18 is 18.
k=\frac{18±12\sqrt{7}}{-6}
Multiply 2 times -3.
k=\frac{12\sqrt{7}+18}{-6}
Now solve the equation k=\frac{18±12\sqrt{7}}{-6} when ± is plus. Add 18 to 12\sqrt{7}.
k=-2\sqrt{7}-3
Divide 18+12\sqrt{7} by -6.
k=\frac{18-12\sqrt{7}}{-6}
Now solve the equation k=\frac{18±12\sqrt{7}}{-6} when ± is minus. Subtract 12\sqrt{7} from 18.
k=2\sqrt{7}-3
Divide 18-12\sqrt{7} by -6.
k=-2\sqrt{7}-3 k=2\sqrt{7}-3
The equation is now solved.
-3k^{2}-18k+57=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.
-3k^{2}-18k+57-57=-57
Subtract 57 from both sides of the equation.
-3k^{2}-18k=-57
Subtracting 57 from itself leaves 0.
\frac{-3k^{2}-18k}{-3}=-\frac{57}{-3}
Divide both sides by -3.
k^{2}+\left(-\frac{18}{-3}\right)k=-\frac{57}{-3}
Dividing by -3 undoes the multiplication by -3.
k^{2}+6k=-\frac{57}{-3}
Divide -18 by -3.
k^{2}+6k=19
Divide -57 by -3.
k^{2}+6k+3^{2}=19+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.
k^{2}+6k+9=19+9
Square 3.
k^{2}+6k+9=28
Add 19 to 9.
\left(k+3\right)^{2}=28
Factor k^{2}+6k+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(k+3\right)^{2}}=\sqrt{28}
Take the square root of both sides of the equation.
k+3=2\sqrt{7} k+3=-2\sqrt{7}
Simplify.
k=2\sqrt{7}-3 k=-2\sqrt{7}-3
Subtract 3 from both sides of the equation.
x ^ 2 +6x -19 = 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 = -19
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) = -19
To solve for unknown quantity u, substitute these in the product equation rs = -19
9 - u^2 = -19
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -19-9 = -28
Simplify the expression by subtracting 9 on both sides
u^2 = 28 u = \pm\sqrt{28} = \pm \sqrt{28}
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{28} = -8.292 s = -3 + \sqrt{28} = 2.292
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.