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3x^{2}-36x-274=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{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 3\left(-274\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -36 for b, and -274 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-36\right)±\sqrt{1296-4\times 3\left(-274\right)}}{2\times 3}
Square -36.
x=\frac{-\left(-36\right)±\sqrt{1296-12\left(-274\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-36\right)±\sqrt{1296+3288}}{2\times 3}
Multiply -12 times -274.
x=\frac{-\left(-36\right)±\sqrt{4584}}{2\times 3}
Add 1296 to 3288.
x=\frac{-\left(-36\right)±2\sqrt{1146}}{2\times 3}
Take the square root of 4584.
x=\frac{36±2\sqrt{1146}}{2\times 3}
The opposite of -36 is 36.
x=\frac{36±2\sqrt{1146}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{1146}+36}{6}
Now solve the equation x=\frac{36±2\sqrt{1146}}{6} when ± is plus. Add 36 to 2\sqrt{1146}.
x=\frac{\sqrt{1146}}{3}+6
Divide 36+2\sqrt{1146} by 6.
x=\frac{36-2\sqrt{1146}}{6}
Now solve the equation x=\frac{36±2\sqrt{1146}}{6} when ± is minus. Subtract 2\sqrt{1146} from 36.
x=-\frac{\sqrt{1146}}{3}+6
Divide 36-2\sqrt{1146} by 6.
x=\frac{\sqrt{1146}}{3}+6 x=-\frac{\sqrt{1146}}{3}+6
The equation is now solved.
3x^{2}-36x-274=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.
3x^{2}-36x-274-\left(-274\right)=-\left(-274\right)
Add 274 to both sides of the equation.
3x^{2}-36x=-\left(-274\right)
Subtracting -274 from itself leaves 0.
3x^{2}-36x=274
Subtract -274 from 0.
\frac{3x^{2}-36x}{3}=\frac{274}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{36}{3}\right)x=\frac{274}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-12x=\frac{274}{3}
Divide -36 by 3.
x^{2}-12x+\left(-6\right)^{2}=\frac{274}{3}+\left(-6\right)^{2}
Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-12x+36=\frac{274}{3}+36
Square -6.
x^{2}-12x+36=\frac{382}{3}
Add \frac{274}{3} to 36.
\left(x-6\right)^{2}=\frac{382}{3}
Factor x^{2}-12x+36. 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-6\right)^{2}}=\sqrt{\frac{382}{3}}
Take the square root of both sides of the equation.
x-6=\frac{\sqrt{1146}}{3} x-6=-\frac{\sqrt{1146}}{3}
Simplify.
x=\frac{\sqrt{1146}}{3}+6 x=-\frac{\sqrt{1146}}{3}+6
Add 6 to both sides of the equation.
x ^ 2 -12x -\frac{274}{3} = 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 = 12 rs = -\frac{274}{3}
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 = 6 - u s = 6 + u
Two numbers r and s sum up to 12 exactly when the average of the two numbers is \frac{1}{2}*12 = 6. 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>
(6 - u) (6 + u) = -\frac{274}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{274}{3}
36 - u^2 = -\frac{274}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{274}{3}-36 = -\frac{382}{3}
Simplify the expression by subtracting 36 on both sides
u^2 = \frac{382}{3} u = \pm\sqrt{\frac{382}{3}} = \pm \frac{\sqrt{382}}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =6 - \frac{\sqrt{382}}{\sqrt{3}} = -5.284 s = 6 + \frac{\sqrt{382}}{\sqrt{3}} = 17.284
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.