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2x^{2}+36x+5=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{-36±\sqrt{36^{2}-4\times 2\times 5}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 36 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-36±\sqrt{1296-4\times 2\times 5}}{2\times 2}
Square 36.
x=\frac{-36±\sqrt{1296-8\times 5}}{2\times 2}
Multiply -4 times 2.
x=\frac{-36±\sqrt{1296-40}}{2\times 2}
Multiply -8 times 5.
x=\frac{-36±\sqrt{1256}}{2\times 2}
Add 1296 to -40.
x=\frac{-36±2\sqrt{314}}{2\times 2}
Take the square root of 1256.
x=\frac{-36±2\sqrt{314}}{4}
Multiply 2 times 2.
x=\frac{2\sqrt{314}-36}{4}
Now solve the equation x=\frac{-36±2\sqrt{314}}{4} when ± is plus. Add -36 to 2\sqrt{314}.
x=\frac{\sqrt{314}}{2}-9
Divide -36+2\sqrt{314} by 4.
x=\frac{-2\sqrt{314}-36}{4}
Now solve the equation x=\frac{-36±2\sqrt{314}}{4} when ± is minus. Subtract 2\sqrt{314} from -36.
x=-\frac{\sqrt{314}}{2}-9
Divide -36-2\sqrt{314} by 4.
x=\frac{\sqrt{314}}{2}-9 x=-\frac{\sqrt{314}}{2}-9
The equation is now solved.
2x^{2}+36x+5=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.
2x^{2}+36x+5-5=-5
Subtract 5 from both sides of the equation.
2x^{2}+36x=-5
Subtracting 5 from itself leaves 0.
\frac{2x^{2}+36x}{2}=-\frac{5}{2}
Divide both sides by 2.
x^{2}+\frac{36}{2}x=-\frac{5}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+18x=-\frac{5}{2}
Divide 36 by 2.
x^{2}+18x+9^{2}=-\frac{5}{2}+9^{2}
Divide 18, the coefficient of the x term, by 2 to get 9. Then add the square of 9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+18x+81=-\frac{5}{2}+81
Square 9.
x^{2}+18x+81=\frac{157}{2}
Add -\frac{5}{2} to 81.
\left(x+9\right)^{2}=\frac{157}{2}
Factor x^{2}+18x+81. 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+9\right)^{2}}=\sqrt{\frac{157}{2}}
Take the square root of both sides of the equation.
x+9=\frac{\sqrt{314}}{2} x+9=-\frac{\sqrt{314}}{2}
Simplify.
x=\frac{\sqrt{314}}{2}-9 x=-\frac{\sqrt{314}}{2}-9
Subtract 9 from both sides of the equation.
x ^ 2 +18x +\frac{5}{2} = 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 2
r + s = -18 rs = \frac{5}{2}
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 = -9 - u s = -9 + u
Two numbers r and s sum up to -18 exactly when the average of the two numbers is \frac{1}{2}*-18 = -9. 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>
(-9 - u) (-9 + u) = \frac{5}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{2}
81 - u^2 = \frac{5}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{2}-81 = -\frac{157}{2}
Simplify the expression by subtracting 81 on both sides
u^2 = \frac{157}{2} u = \pm\sqrt{\frac{157}{2}} = \pm \frac{\sqrt{157}}{\sqrt{2}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-9 - \frac{\sqrt{157}}{\sqrt{2}} = -17.860 s = -9 + \frac{\sqrt{157}}{\sqrt{2}} = -0.140
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.